Exponential and Logarithmic Functions — Free Algebra 2 Review Games.

10^2 = 100, so log_10(100) = 2.

2^5 = 2*2*2*2*2 = 32.

log_b(x) = y means b^y = x.

2^3 = 8, so log_2(8) = 3.

Any nonzero base to the power 0 equals 1.

Product rule of logarithms: log(x) + log(y) = log(xy).

Power rule: \(\log(x^3) = 3\log(x)\).

81 = 3^4, so x = 4.

After 15 years = 3 doublings: 100 * 2^3 = 800.

ln(e) = log_e(e) = 1.

3^2 = x-1, 9 = x-1, x = 10.

16 = 2^4, so x+1 = 4, x = 3.

Quotient rule: log(x) - log(y) = log(x/y).

1000 * (1.05)^10 = 1000 * 1.62889 = $1628.89.

Take ln: 2x = ln(7), x = ln(7)/2.

log_10(1000) asks: 10 to what power equals 1000? Since 10^3 = 1000, the answer is 3. A common mistake is confusing the number of zeros (3) with the result itself — the number of zeros in 1000 does equal the exponent here, but only because we are using base 10.

log_b(1) asks: b to what power equals 1? Since any nonzero base raised to the power of 0 equals 1 (b^0 = 1), the answer is always 0. Choice 1 is incorrect because b^1 = b, not 1.

ln(1) is the natural logarithm of 1, which asks: e to what power equals 1? Since e^0 = 1, ln(1) = 0. A common error is confusing ln(e) = 1 with ln(1) = 1 — these are different inputs.

log_b(b) asks: b to what power equals b? Since b^1 = b, the answer is always 1. This is a fundamental logarithm identity. Choice 0 is wrong because log_b(b) = 0 would mean b^0 = b, which is false (b^0 = 1).

When the base is between 0 and 1, the exponential function models decay — as x increases, y gets smaller and smaller, approaching but never reaching 0. This makes the graph decreasing. Choice A is incorrect because decreasing functions approach 0 as x increases, not increasing ones.

log_5(125) asks: 5 to what power equals 125? Since 5^3 = 125, the answer is 3. Choice 2 is a distractor because 5^2 = 25, not 125.

The conversion rule is: b^y = x is equivalent to log_b(x) = y. The base of the exponent (b) becomes the base of the logarithm, the exponent (y) becomes the result, and the value (x) becomes the argument. Choice D incorrectly swaps x and y inside the logarithm.

Evaluate each term separately: log_3(27) = 3 (since 3^3 = 27) and log_3(3) = 1 (since 3^1 = 3). Adding gives 3 + 1 = 4. Choice D is wrong because the product rule log_b(M) + log_b(N) = log_b(MN) gives log_3(81) = 4, not log_3(30) — you multiply the arguments, not add them.

Write both sides as powers of the same base: 64 = 4^3, so 4^x = 4^3, which gives x = 3. Choice A is incorrect because 4^2 = 16, not 64. Choice C is wrong because 4^4 = 256.

The change-of-base formula states log_b(x) = log(x) / log(b), where the logarithm in numerator and denominator can be any common base (base 10 or natural log). Choice A inverts the fraction — it would equal log_x(b), not log_b(x).

Using exponential growth: A = 200(1.30)^2 = 200 times 1.69 = 338. Choice A (260) incorrectly applies simple interest: 200 + 200(0.30)(2) = 260. Exponential growth compounds at each step, so the second hour grows from 260, not 200.

Use the power rule first: \(2\log(x) = \log(x^2)\) and \(3\log(y) = \log(y^3)\). Then apply the product rule: \(\log(x^2) + \log(y^3) = \log(x^2 y^3)\). Choice A incorrectly treats logarithm coefficients as multipliers of the arguments rather than exponents.

The argument of a logarithm must be strictly positive: x - 4 > 0, so x > 4. Choice A (x > 0) is the domain of log_2(x) without the shift — the horizontal shift of 4 units right moves the boundary from 0 to 4.

Convert to exponential form: log_4(x) = 3 means 4^3 = x. Since 4^3 = 64, x = 64. Choice C (16) results from computing 4^2 = 16 — a common off-by-one error when evaluating powers.

f(-2) = 5^(-2) = 1/(5^2) = 1/25. A negative exponent means the reciprocal of the positive power. Choice A (-25) is a common error that incorrectly negates the result instead of taking the reciprocal.

Apply the product rule to the numerator: \(\log(x^2 y) = \log(x^2) + \log(y) = 2\log(x) + \log(y)\). Apply the quotient rule for division by z: subtract \(\log(z)\). Result: \(2\log(x) + \log(y) - \log(z)\). Choice A incorrectly subtracts \(\log(y)\) instead of adding it.

Use the decay formula A = 200(0.85)^3. Computing: (0.85)^3 = 0.614125, so A = 200 times 0.614 = 122.8 g. Choice B (127.5 g) uses simple decay: 200 - 200(0.15)(3) = 110 g — wait, that gives 110, not 127.5. Choice B is 200*(0.85)^2 ≈ 144.5 ... actually 200*0.85 = 170, 170*0.85 = 144.5, 144.5*0.85 = 122.8. The distractors include 200 - 200*0.15*3 = 110 and 200*(1-0.45) = 110 as well. 127.5 comes from 200*(1 - 3*0.15/2) which is a wrong approximation.

Exponential decay occurs when the base is between 0 and 1 (exclusive). In y = (2/3)^x, the base 2/3 is between 0 and 1, so y decreases as x increases. Choices A, B, and C all have bases greater than 1 (2, 1.5, and e respectively), so they all model exponential growth.

Apply the product rule: \(\log_2(x(x - 6)) = 4\). Convert to exponential form: \(x(x - 6) = 2^4 = 16\). Expand: \(x^2 - 6x - 16 = 0\). Factor: \((x - 8)(x + 2) = 0\), giving \(x = 8\) or \(x = -2\). Check the domain: both \(x\) and \(x - 6\) must be positive, so \(x\) must be greater than 6. \(x = -2\) is rejected. The only solution is \(x = 8\).

Rewrite both sides using base 3: 9 = 3^2 and 27 = 3^3. So the equation becomes 3^(2x) = 3^(3(x - 1)). Since the bases are equal, set the exponents equal: 2x = 3x - 3. Solving: -x = -3, so x = 3. Choice A is wrong because 9^1 = 9 and 27^0 = 1, which are not equal.

Factor 18 = 2 times 3^2. Then log(18) = log(2) + log(3^2) = log(2) + 2log(3) = 0.301 + 2(0.477) = 0.301 + 0.954 = 1.255. Choice C (0.778) is just log(2) + log(3) = log(6), which incorrectly ignores that 18 = 2 times 9, not 2 times 3.

Let u = 3^x, so 3^(2x) = u^2. The equation becomes u^2 - 12u + 27 = 0. Factor: (u - 3)(u - 9) = 0, giving u = 3 or u = 9. Back-substitute: 3^x = 3 gives x = 1, and 3^x = 9 = 3^2 gives x = 2. Both solutions are valid. Choice D lists the u values, not the x values — a common substitution error.

Set 8 = 2^(t/12). Since 8 = 2^3, we have 2^3 = 2^(t/12), so t/12 = 3 and t = 36 years. Intuitively: the investment doubles at year 12, doubles again (4x) at year 24, and doubles a third time (8x) at year 36. Choice A (24 years) only achieves a 4x increase.

Apply the product rule: \(\log_3(x(x + 6)) = 3\). Convert: \(x(x + 6) = 3^3 = 27\). Expand: \(x^2 + 6x - 27 = 0\). Factor: \((x + 9)(x - 3) = 0\), giving \(x = -9\) or \(x = 3\). Since \(\log_3(x)\) requires \(x\) to be greater than 0, \(x = -9\) is rejected (undefined). The only valid solution is \(x = 3\). Choice B lists both algebraic results without applying the domain restriction.

The half-life occurs when A = A_0/2, so 1/2 = e^(-0.0347t). Take the natural log of both sides: ln(1/2) = -0.0347t, which gives -ln(2) = -0.0347t. Solving: t = ln(2)/0.0347 = 0.6931/0.0347 = approximately 20 years. Choice D (30 years) would correspond to a decay constant of about 0.0231, not 0.0347.

The exponential form b^n = x is equivalent to log_b(x) = n. Therefore 4^3 = 64 becomes log_4(64) = 3. Choice A incorrectly places the exponent (3) as the base of the logarithm rather than keeping 4 as the base.

log_3(81) asks '3 to what power gives 81?' Since 3^4 = 81, the answer is 4. Choice B is wrong because 3^3 = 27, not 81. Choice D confuses the argument of the logarithm with the answer.

The y-intercept occurs at x = 0. f(0) = 7^0 = 1. Any nonzero base raised to the power 0 equals 1, so all exponential functions of the form b^x pass through the point (0, 1). Choice B confuses the base value with the y-intercept.

By the inverse property of logarithms, ln(e^x) = x for all real x. So ln(e^8) = 8. The natural logarithm and the exponential function with base e are inverses of each other, so they undo each other completely. Choice A incorrectly multiplies 8 by e rather than recognizing the inverse relationship.

The Quotient Property of logarithms states that the log of a quotient equals the difference of the logs of the numerator and denominator. The Product Property applies to multiplication (yielding a sum of logs), and the Power Property applies to exponents (yielding a coefficient times a log).

For the quantity to be halved every 5 years, the exponent must be t/5. When t = 5, Q = 100 * (0.5)^(5/5) = 100 * 0.5 = 50, which is correct. Choice A would halve the quantity every single year rather than every 5 years. Choice D is linear, not exponential.

Since 0.01 = 10^(-2), we have log_10(0.01) = log_10(10^(-2)) = -2. The exponent to which 10 must be raised to produce 0.01 is -2. Choice A corresponds to log_10(0.1) = -1, which is one power of 10 larger. Choice D is positive, but the log of a number between 0 and 1 must be negative.

Exponential functions of the form f(x) = b^x, where b > 0 and b not equal to 1, are defined for all real values of x. The exponent can be any real number. Unlike logarithmic functions, there is no restriction on the input. Choice A incorrectly applies the domain restriction of a logarithmic function to an exponential function.

Rewrite 64 as a power of 4: 64 = 4^3. The equation becomes 4^(x-1) = 4^3, so x - 1 = 3, giving x = 4. Verify: 4^(4-1) = 4^3 = 64. Choice B gives x = 3, meaning 4^(3-1) = 4^2 = 16, which does not equal 64.

The change of base formula states log_b(a) = log(a) / log(b). Applying this gives log_8(50) = log(50) / log(8). Choice A inverts the fraction, which would give log_50(8) instead. Choice D incorrectly applies the Quotient Property, which would simplify log(50/8), not change the base of a logarithm.

Continuous exponential growth follows the formula A(t) = A_0 * e^(rt), where r is the growth rate as a decimal. With A_0 = 400 and r = 0.06, the model is A(t) = 400 * e^(0.06t). Choice A models discrete annual growth, not continuous growth. Choice D incorrectly uses r = 6 instead of the decimal 0.06.

First apply the Power Property: \(3\log(x) = \log(x^3)\). Then apply the Quotient Property: \(\log(x^3) - \log(y) = \log(x^3 / y)\). Choice B incorrectly treats the coefficient 3 as a multiplier of x inside the argument rather than as an exponent. Choice C uses multiplication instead of division.

To find the inverse of \(y = 2^x\), swap x and y to get \(x = 2^y\), then convert to logarithmic form: \(y = \log_2(x)\). So \(f^{-1}(x) = \log_2(x)\). The inverse of an exponential function is the logarithm with the same base. Choice D produces a reflection across the y-axis, not the inverse function.

Take the common logarithm of both sides: x * log(6) = log(200). Solve for x: x = log(200) / log(6) ≈ 2.301 / 0.778 ≈ 2.96. Choice B results from dividing 200 by 6 arithmetically rather than applying the change of base method. Choice C comes from dividing log(6) / log(200), which inverts the fraction.

Rewrite the square root as a 1/2 power: \((x^5 / y^2)^{1/2} = x^{5/2} / y\). Then \(\log_b(x^{5/2} / y) = (5/2)\log_b(x) - \log_b(y)\). Choice B forgets to apply the 1/2 exponent from the square root to the entire expression. Choice D applies the square root to \(x^5\) correctly but not to \(y^2\), leaving \(y^2\) unchanged.

Since 27 = 3^3, apply the Power Property: log_b(27) = log_b(3^3) = 3 * log_b(3) = 3 * 0.68 = 2.04. Choice B gives 2 * 0.68 = 1.36, which equals log_b(9) since 9 = 3^2. Choice A divides 0.68 by 3 rather than multiplying, which would give log_b(3^(1/3)).

The parent function 2^x has a horizontal asymptote at y = 0. Multiplying by 3 stretches the graph vertically but does not affect the asymptote. Subtracting 5 shifts the entire graph down by 5 units, moving the horizontal asymptote from y = 0 to y = -5. Choice D is the asymptote of the untransformed function 2^x before the vertical shift is applied.

Take the common log of both sides: (x + 3)log(2) = x * log(5). Expand: x*log(2) + 3log(2) = x*log(5). Collect x terms: 3log(2) = x*(log(5) - log(2)). Divide: x = 3log(2) / (log(5) - log(2)) ≈ 2.27. Choice B adds the logs in the denominator, an error that does not follow from the algebra. Choice D uses only log(5) in the denominator, omitting the subtraction of log(2).

Use the continuous compound interest formula A = Pe^(rt). Substitute: 2500 = 1000 * e^(0.05t). Divide both sides by 1000: 2.5 = e^(0.05t). Take the natural log: ln(2.5) = 0.05t. Solve: t = ln(2.5) / 0.05 ≈ 0.916 / 0.05 ≈ 18.3 years. Choice A corresponds approximately to doubling to $2,000 rather than reaching $2,500.

Apply the Quotient Property: log_5((x + 2) / (x - 2)) = 1. Convert to exponential form: (x + 2) / (x - 2) = 5. Cross-multiply: x + 2 = 5(x - 2) = 5x - 10. Solve: 12 = 4x, so x = 3. Verify the domain: x + 2 = 5 > 0 and x - 2 = 1 > 0, so x = 3 is valid. Choice C (x = 2) makes x - 2 = 0, which creates an undefined logarithm.

Substitute u = e^x to obtain the quadratic u^2 - 7u + 12 = 0. Factor: (u - 3)(u - 4) = 0, so u = 3 or u = 4. Solve e^x = 3 to get x = ln(3), and e^x = 4 to get x = ln(4). Both solutions are valid because e^x > 0 for all x, so neither is extraneous. Choice B incorrectly treats the values of u (3 and 4) as the final answers instead of taking the natural log.

Factor 72 as 2^3 * 3^2. Then log(72) = log(2^3 * 3^2) = 3log(2) + 2log(3) = 3(0.301) + 2(0.477) = 0.903 + 0.954 = 1.857. Choice A equals log(36) since 2log(2) + 2log(3) = 0.602 + 0.954 = 1.556. Choice D equals log(6) = log(2) + log(3) = 0.778, using only one factor of each prime.

Use the half-life model A(t) = A_0 * (1/2)^(t/h). Set up 50 = 800 * (1/2)^(t/20). Simplify: 1/16 = (1/2)^(t/20). Since (1/2)^4 = 1/16, the equation gives t/20 = 4, so t = 80 years. Choice A (40 years) represents only 2 half-lives: 800 to 400 to 200 grams, leaving far more than 50 grams.

Apply the change of base formula using base 2: log_4(9) = log_2(9) / log_2(4). Since log_2(9) = log_2(3^2) = 2log_2(3) = 2p and log_2(4) = log_2(2^2) = 2, we get log_4(9) = 2p / 2 = p. Choice C (2p) is a common error where students compute 2log_2(3) = 2p but forget to divide by log_2(4) = 2.

Rewrite 3^(x+1) = 3 * 3^x. Substitute: 2(3)(3^x) - 5(3^x) = 9, which gives 6(3^x) - 5(3^x) = 9, so 3^x = 9 = 3^2, and x = 2. Verify: 2 * 3^3 - 5 * 3^2 = 54 - 45 = 9. Choice A (x = 1) gives 2 * 3^2 - 5 * 3^1 = 18 - 15 = 3, not 9. The key step is factoring out 3^x after expanding 3^(x+1).