Exercises
Explore the essential relationships between exponential and logarithmic functions. This quiz covers conversion between forms, logarithm laws, domains, equations, inverse functions, graph behavior, transformations, asymptotes, growth models, doubling time, and scientific applications. Questions range from foundational calculations to multi-step reasoning and visual graph interpretation.
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The equivalence b^y = x means log_b(x) = y. Therefore, 2^5 = 32 is equivalent to log_2(32) = 5.
Because 3^4 = 81, the exponent needed to produce 81 from base 3 is 4. Thus, log_3(81) = 4.
A logarithm requires a positive argument. Since x - 4 > 0, the domain is x > 4. The graph exists only to the right of its vertical asymptote x = 4.
The product rule separates multiplication, the quotient rule creates subtraction, and the power rule moves exponents forward. Since √z = z^(1/2), its term is -(1/2)log_b(z).
The power rule gives 2ln(x) = ln(x^2). The quotient rule then gives ln(x^2) - ln(y) = ln(x^2/y).
Write 125 as 5^3. Equal bases imply x - 1 = 3, so x = 4.
Convert to exponential form: x + 1 = 2^3 = 8. Subtracting 1 gives x = 7, which satisfies the logarithm's domain.
Combine the logarithms: ln(x(x - 3)) = ln(4), so x^2 - 3x - 4 = 0. The roots are 4 and -1, but the original expressions require x > 3. Therefore, only x = 4 is valid.
Logarithm laws apply to products, quotients, and powers, but there is no corresponding rule for splitting a sum. In general, log_b(M + N) does not equal log_b(M) + log_b(N).
Reflection across y = x produces an inverse function. The inverse of y = 3^x is y = log_3(x).
An exponential function with base between 0 and 1 decreases. At x = 0 its value is 1, and its horizontal asymptote is y = 0.
Replacing x with x - 3 shifts the graph 3 units right. Adding 1 outside the exponential shifts it 1 unit up.
The logarithm's argument approaches zero when x + 5 = 0. Therefore, the graph has the vertical asymptote x = -5. The -2 creates only a vertical shift.
The initial value is 200. The one-period growth factor is 212/200 = 1.06, representing 6% growth. Thus, P(t) = 200(1.06)^t.
At doubling, 2A_0 = A_0e^(0.08t). Dividing by A_0 and taking natural logs gives ln(2) = 0.08t, so t = ln(2)/0.08.
Since log_10(10^-5) = -5, the leading negative sign gives pH = -(-5) = 5.
Taking natural logs gives xln(4) = ln(10). Dividing by ln(4) yields x = ln(10)/ln(4), which is also log_4(10).
Combine the logs to get log_10(x(x - 9)) = 1, so x(x - 9) = 10. Factoring x^2 - 9x - 10 gives roots 10 and -1. The domain requires x > 9, leaving x = 10.
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