Exponential and Logarithmic Functions — Free Pre-Calculus Review Games.

ln(1) = 0 because e^0 = 1.

10^3 = 1000, so log(1000) = 3.

e is approximately 2.71828.

f(0) = e^0 = 1.

b^1 = b, so log_b(b) = 1.

\(\ln(x^2 y) = \ln(x^2)+\ln(y) = 2\ln(x)+\ln(y)\).

2^5 = 32, so x = 32.

Sum becomes product, difference becomes quotient: log(3x/2).

Take natural log: x = ln(20).

(1/2)^(15/5) = (1/2)^3 = 1/8.

\(\ln((x-1)(x+1)) = \ln(3)\), \(x^2-1=3\), \(x^2=4\), \(x=2\) (reject \(-2\) since \(\ln\) domain).

125 = 5^3, so 2x-1=3, 2x=4, x=2.

Change of base: log_a(b) = ln(b)/ln(a).

A = 1000*e^(0.3) = 1000*1.34986 = $1349.86.

\(x^2 = 10^4 = 10000\), \(x = \pm 100\).

The natural logarithm is only defined for positive inputs because \(\ln(x)\) answers the question \(e^y = x\), and \(e^y\) is always positive. The choice "All real numbers" is wrong because negative numbers and zero have no exponent \(y\) that makes \(e^y\) equal them. Always check that the argument of any logarithm is strictly positive before solving.

By definition, \(\log_b(1) = 0\) because \(b^0 = 1\) for any valid base \(b\). The distractor "\(1\)" is wrong since that would require \(b^1 = 1\), which is only true if \(b=1\), and \(1\) is not a valid logarithmic base. Remember that any log of \(1\) always equals zero regardless of the base.

Any nonzero number raised to the power of zero equals \(1\), and \(e^0 = 1\) follows this exponent rule. The distractor "\(e\)" confuses \(e^0\) with \(e^1\), which actually equals \(e\). This zero-exponent rule applies universally across all exponential functions.

Since \(\ln(x)\) and \(e^x\) are inverse functions, \(\ln(e) = \ln(e^1) = 1\). The distractor "\(e\)" mistakenly treats the natural log as if it returns the base itself rather than the exponent. Whenever the argument of \(\ln\) matches \(e\) raised to a power, the answer is simply that power.

\(\log_2(8) = 3\) because \(2^3 = 8\), matching the definition of a logarithm as an exponent. The distractor "\(4\)" would require \(2^4 = 16\), which is not equal to \(8\). To evaluate a log by hand, rewrite it as an exponential equation and solve for the exponent.

When the base \(b\) is greater than \(1\), each unit increase in \(x\) multiplies the output by a factor greater than \(1\), producing growth. The distractor "\(0 < b < 1\)" actually describes decay, since repeated multiplication by a fraction shrinks the value. Always check whether the base is above or below \(1\) to classify a model as growth or decay.

Since \(\log_b(1) = 0\) for any valid base, every logarithmic graph of this form crosses the x-axis at \((1,0)\). The distractor "\((0,1)\)" is wrong because \(x=0\) is not even in the domain of a logarithm. This fixed point is a quick way to sketch or verify any log graph.

Because \(e^x\) is always positive and gets arbitrarily close to but never reaches zero, the range is \(y > 0\). The distractor "All real numbers" describes the domain of \(e^x\), not its output values. Exponential functions with a positive base never produce zero or negative outputs.

\(\log(100) = 2\) because the common logarithm has an implied base of \(10\), and \(10^2 = 100\). The distractor "\(10\)" confuses the base of the logarithm with its actual value. Whenever no base is written, assume base \(10\) for \(\log\).

The natural logarithm is defined specifically as \(\log_e(x)\), so its base is the irrational constant \(e \approx 2.718\). The distractor "\(10\)" describes the common logarithm, written as \(\log(x)\), not \(\ln(x)\). Distinguishing \(\ln\) from \(\log\) by their bases is essential before applying log rules.

Since the base \(2\) is greater than \(1\), larger values of \(x\) produce larger outputs, making the function increasing. The distractor "Decreasing" would apply if the base were between \(0\) and \(1\). Comparing the base to \(1\) instantly tells you the direction of an exponential function's growth.

\(\log_5(25) = 2\) because \(5^2 = 25\), so the exponent that produces \(25\) from base \(5\) is \(2\). The distractor "\(5\)" mistakes the base for the answer rather than solving for the exponent. Rewriting logarithms as exponential equations makes this type of evaluation straightforward.

As \(x \to -\infty\), \(3^x\) approaches but never reaches \(0\), making \(y=0\) the horizontal asymptote. The distractor "\(y=3\)" mistakes the base of the function for its asymptote. All basic exponential functions of the form \(b^x\) share the asymptote \(y=0\) unless the function includes a vertical shift.

The quotient rule splits the log into a difference, and the power rule brings the exponent \(3\) down as a coefficient on \(\ln(y)\), giving \(\ln(x) - 3\ln(y)\). The distractor "\(\ln(x) + 3\ln(y)\)" incorrectly applies addition instead of subtraction for a quotient. Always apply the quotient rule before the power rule when expanding a log of a fraction with an exponent.

The power rule turns \(2\ln(x)\) into \(\ln(x^2)\), and the quotient rule combines the subtraction into \(\ln\left(\frac{x^2}{y}\right)\). The distractor "\(\ln(x^2 - y)\)" incorrectly treats logarithm subtraction as subtraction inside the argument rather than division. When condensing, coefficients become exponents first, then addition or subtraction of logs becomes multiplication or division inside one log.

Rewriting in exponential form gives \(x + 2 = 3^4 = 81\), so \(x = 79\). The distractor "\(x = 81\)" forgets to subtract the \(2\) after converting to exponential form. Always convert a log equation to its exponential equivalent, then solve the resulting algebraic equation for the variable.

Exponentiating both sides with base \(e\) gives \(x = e^2 \approx 7.39\). The distractor "\(0.69\)" is actually \(\ln(2)\), confusing the operation direction. To undo a natural log, raise \(e\) to the power equal to both sides of the equation.

Since \(t/h = 24/8 = 3\) half-lives have passed, the remaining fraction is \((0.5)^3 = \dfrac{1}{8}\). The distractor "\(\dfrac{1}{4}\)" corresponds to only \(2\) half-lives, not \(3\). Always compute the number of half-lives first by dividing elapsed time by the half-life before applying the decay factor.

Since \(16 = 2^4\), the equation becomes \(2^{x+1} = 2^4\), so setting exponents equal gives \(x + 1 = 4\) and \(x = 3\). The distractor "\(x = 4\)" forgets to subtract \(1\) after matching exponents. When both sides can be written with the same base, equate the exponents directly to solve for the variable.

Applying the formula, \(\text{pH} = -\log(10^{-5}) = -(-5) = 5\), using the log rule \(\log(10^n) = n\). The distractor "\(-5\)" forgets to apply the negative sign in the pH formula. Recognizing \(\log_{10}\) of a power of ten simply returns the exponent makes these calculations quick.

Taking \(\log\) of both sides gives \(x = \dfrac{\log(50)}{\log(3)} \approx 3.56\). The distractor "\(16.67\)" comes from dividing \(50\) by \(3\) instead of using logarithms to solve for the exponent. When the variable is in the exponent and the bases can't be matched, take the log of both sides and use the quotient of logs to isolate \(x\).

Substituting gives \(A = 500e^{0.4} \approx 500(1.4918) \approx \\)746$. The distractor "\(\\)700$" results from a rounding or arithmetic shortcut that ignores the actual value of \(e^{0.4}\). Continuous compounding always uses the constant \(e\) raised to the product of the rate and time.

Combining logs gives \(\log(x(x-3)) = 1\), so \(x^2 - 3x - 10 = 0\), factoring to \(x = 5\) or \(x = -2\); but \(x = -2\) makes \(\log(x)\) undefined, so only \(x = 5\) works. The distractor "\(x = 5\) or \(x = -2\)" fails to reject the extraneous negative solution. Whenever solving log equations by combining terms, check every solution against the original domain restrictions.

Setting \(2P_0 = P_0 e^{0.03t}\) and solving gives \(t = \dfrac{\ln(2)}{0.03} \approx 23.1\) years. The distractor "\(33.3\) years" incorrectly computes \(\dfrac{1}{0.03}\) instead of dividing \(\ln(2)\) by the growth rate. Doubling-time problems always require solving for \(t\) using \(\ln(2)\) divided by the growth constant.

Taking the natural log of both sides gives \(2x = \ln(7)\), so \(x = \dfrac{\ln(7)}{2} \approx 0.97\). The distractor "\(1.95\)" forgets to divide by \(2\) after taking the log. When the exponent contains a coefficient on \(x\), isolate \(x\) by dividing after applying the logarithm.

By the change of base formula, \(\log_5(12) = \dfrac{\ln(12)}{\ln(5)} \approx \dfrac{2.485}{1.609} \approx 1.54\). The distractor "\(2.40\)" incorrectly divides \(12\) by \(5\) instead of using logarithms. The change of base formula lets you evaluate any logarithm using a calculator's built-in \(\ln\) or \(\log\) function.

Combining logs gives \(\ln\left(\dfrac{x+3}{x}\right) = 1\), so \(\dfrac{x+3}{x} = e\), and solving yields \(x = \dfrac{3}{e-1} \approx 1.75\). The distractor "\(3.00\)" comes from mistakenly setting \(x+3 = ex\) without properly isolating \(x\) across the equation. When condensing a log difference into a quotient, exponentiate both sides carefully and solve the resulting rational equation for the variable.

Subtracting a constant from \(x\) inside the function shifts the graph horizontally to the right by that amount, so \(y = \ln(x-2)\) moves the parent graph right 2 units. The distractor "Shifted left 2 units" confuses the sign convention for horizontal shifts. Inside-the-function changes to \(x\) always shift graphs horizontally in the opposite direction of the sign.

Rewriting with base \(2\) gives \(2^{2x} = 2^{3(x-1)}\), so \(2x = 3x - 3\), which solves to \(x = 3\). The distractor "\(x = 1\)" results from an arithmetic error when distributing the \(3\) across \((x-1)\). When bases share a common factor, express both sides using that common base before equating exponents.

Converting to exponential form gives \(x - 1 = 2^3 = 8\), so \(x = 9\). The distractor "\(x = 8\)" stops after finding \(2^3\) but forgets to add \(1\) back to isolate \(x\). Always finish converting the exponential equation into a fully solved linear equation for the variable.

Setting \(0.5 = e^{-6k}\) and solving gives \(k = \dfrac{\ln(2)}{6} \approx 0.1155\). The distractor "\(0.6931\)" is \(\ln(2)\) itself, forgetting to divide by the half-life. Finding a decay constant from a half-life always involves dividing \(\ln(2)\) by the given half-life value.

Exponentiating gives \(2x = e^3 \approx 20.09\), so dividing by \(2\) yields \(x \approx 10.04\). The distractor "\(20.09\)" forgets to divide by \(2\) after solving for \(2x\). Always isolate the full expression inside the logarithm first, then solve for the variable itself.

Combining logs gives \(\log_2(x(x-2)) = 3\), so \(x^2 - 2x - 8 = 0\), factoring to \(x=4\) or \(x=-2\); since \(\log_2(-2)\) is undefined, only \(x=4\) is valid. The distractor "\(x = 4\) or \(x = -2\)" fails to eliminate the extraneous root outside the domain. In any logarithmic equation, every candidate solution must be checked against the original domain before being accepted.

Letting \(u = e^x\) turns the equation into \(u^2 - 5u + 6 = 0\), which factors to \((u-2)(u-3)=0\), giving \(u=2\) or \(u=3\), so \(x = \ln(2)\) or \(x = \ln(3)\). The distractor "\(x = 2\) or \(x = 3\)" forgets to undo the substitution by taking the natural log at the end. Quadratic-in-form exponential equations require a substitution step, and the final answer must return to the original variable using logarithms.

Taking \(\ln\) of both sides gives \(x\ln 3 = (x-1)\ln 5\), which rearranges to \(x(\ln 3 - \ln 5) = -\ln 5\), so \(x = \dfrac{\ln 5}{\ln 5 - \ln 3} \approx 3.15\). The distractor "\(1.46\)" comes from an algebra error that fails to properly distribute \(\ln 5\) across \((x-1)\). When variables appear on both sides as exponents with different bases, take logs of both sides and carefully solve the resulting linear equation for \(x\).

Combining the right side gives \(\ln(x^2-1) = \ln(2(x+1))\), so \(x^2 - 1 = 2x + 2\), which simplifies to \(x^2 - 2x - 3 = 0\), factoring to \(x=3\) or \(x=-3\); but \(x=-3\) makes \(x+1\) negative, so only \(x=3\) satisfies the domain. The distractor "\(x = 3\) or \(x = -3\)" ignores that \(\ln(x+1)\) requires \(x > -1\). Domain restrictions on every log term in an equation must all be satisfied simultaneously by a valid solution.

Solving \(5000 = 2000e^{10r}\) gives \(e^{10r} = 2.5\), so \(r = \dfrac{\ln(2.5)}{10} \approx 0.0916\), or \(9.16\%\). The distractor "\(25.00\%\)" mistakenly uses the ratio \(5000/2000\) directly as a percentage instead of solving for the exponential rate. Continuous compounding problems that ask for a rate require isolating the exponent using the natural log after dividing the two amounts.

Rewriting \(8\) as \(2^3\) gives \(2^{x^2} = 2^{3x}\), so \(x^2 = 3x\), which factors to \(x(x-3)=0\), giving \(x = 0\) or \(x = 3\). The distractor "\(x = 3\) only" discards a valid solution because it overlooks that \(x=0\) also satisfies the original equation. When exponents can be matched to a common base, solve the resulting polynomial equation completely rather than assuming only nonzero roots count.

Combining the left side gives \(\log\left(\dfrac{x+2}{x-1}\right) = \log(4)\), so \(\dfrac{x+2}{x-1} = 4\), which solves to \(x+2 = 4x-4\) and \(x = 2\); checking, both \(x+2\) and \(x-1\) remain positive at \(x=2\). The distractor "\(x = -2\)" would make \(\log(x-1)\) evaluate at a negative number, violating the domain of the logarithm. Any solution to a log equation must keep every original logarithmic argument positive.

Since \(9 = 3^2\), reaching \(9\) times the population is equivalent to tripling twice, so it takes \(2 \times 15 = 30\) years due to the multiplicative nature of exponential growth. The distractor "\(45\) years" incorrectly assumes tripling three times is needed instead of recognizing \(9 = 3^2\). Exponential growth problems involving powers of the same growth factor can often be solved by counting how many growth periods are needed rather than resolving for \(k\) explicitly.

Using the change of base formula, \(\log_4(x) = \dfrac{\log_2(x)}{2}\), so the equation becomes \(\dfrac{\log_2(x)}{2} = \log_2(x) - 1\), which gives \(\log_2(x) = 2\), so \(x = 4\). The distractor "\(x = 16\)" comes from an error solving the linear equation in \(\log_2(x)\) after the substitution. When two logarithms in an equation have different bases that are powers of the same number, rewrite them using a common base before solving.

Multiplying by \(e^x\) gives \(e^{2x} - 3e^x + 1 = 0\); letting \(u = e^x\) and applying the quadratic formula gives \(u = \dfrac{3+\sqrt{5}}{2} \approx 2.618\), so \(x = \ln(2.618) \approx 1.20\). The distractor "\(0.96\)" comes from selecting the smaller root of the quadratic without checking it corresponds to a negative \(x\) value. Equations mixing \(e^x\) and \(e^{-x}\) often require multiplying through by \(e^x\) to convert them into a solvable quadratic in \(u = e^x\).

Setting \(10000 = 25000(0.85)^t\) gives \((0.85)^t = 0.4\), and taking logs yields \(t = \dfrac{\ln(0.4)}{\ln(0.85)} \approx 5.63\) years. The distractor "\(3.75\) years" comes from dividing the dollar amounts directly rather than solving the exponential equation with logarithms. Depreciation problems require isolating the exponential factor and solving for \(t\) using logarithms, just like growth problems.

Rewriting gives \(\ln(x^2) - \ln(x-2) = \ln(8)\), so \(\ln\left(\dfrac{x^2}{x-2}\right) = \ln(8)\), giving \(\dfrac{x^2}{x-2} = 8\), which leads to \(x^2 - 8x + 16 = 0\), factoring to \((x-4)^2 = 0\), so \(x=4\); this satisfies \(x-2>0\) and \(x>0\). The distractor "\(x = 2\)" would make \(\ln(x-2)\) undefined since \(\ln(0)\) does not exist. Repeated roots from a squared factor should still be checked against domain restrictions before being finalized as the solution.

Using \(t = \dfrac{\ln(2)}{r}\), the \(5\%\) investment doubles in \(\dfrac{\ln 2}{0.05} \approx 13.86\) years and the \(6\%\) investment doubles in \(\dfrac{\ln 2}{0.06} \approx 11.55\) years, a difference of about \(2.31\) years. The distractor "\(13.86\) years" gives only the doubling time for the \(5\%\) investment rather than the difference between the two doubling times. Comparing doubling times for different continuous growth rates requires computing each separately with \(\ln(2)/r\) before subtracting.