Exponential and Logarithmic Functions

1. The exponential equation $b^y = x$ is equivalent to the logarithmic equation $\log_b x = y$.

Applying this with $b = 2$, $y = 5$, and $x = 32$:

$\log_2 32 = 5$

2. Rewrite using the definition of logarithms:

$3^x = 81$

Since $81 = 3^4$, we have:

$3^x = 3^4$

Since the bases are equal, the exponents must be equal:

$x = 4$

Verification: $3^4 = 81$ ✓

3. Set $P(t) = 16{,}000$ and solve:

$500 \cdot 2^t = 16{,}000$

$2^t = \frac{16{,}000}{500} = 32$

Rewrite 32 as a power of 2: $32 = 2^5$, so:

$2^t = 2^5 \implies t = 5$

In logarithmic form: $t = \log_2 32 = 5$.

It takes 5 hours for the population to reach 16,000.

Verification: $P(5) = 500 \cdot 2^5 = 500 \cdot 32 = 16{,}000$ ✓

4. Convert from logarithmic to exponential form using $\log_b x = y \iff b^y = x$:

$\log_5(x + 6) = 3 \implies 5^3 = x + 6$

$125 = x + 6$

$x = 119$

Verification: $\log_5(119 + 6) = \log_5(125) = \log_5(5^3) = 3$ ✓

5. (a) To find the inverse, start with $y = 3^x$ and swap $x$ and $y$:

$x = 3^y$

Rewrite in logarithmic form:

$y = \log_3 x$

Therefore $f^{-1}(x) = \log_3 x$.

(b) Compute from the inside out:

$f^{-1}(27) = \log_3 27 = \log_3 3^3 = 3$

$f(f^{-1}(27)) = f(3) = 3^3 = 27 \checkmark$

This confirms that $f$ and $f^{-1}$ are inverses: applying one after the other returns the original input (F-BF.B.5).

(c) Since $f(x) = 3^x$ has domain $(-\infty, \infty)$ and range $(0, \infty)$, its inverse $f^{-1}(x) = \log_3 x$ has:

• Domain: $(0, \infty)$ — you can only take the logarithm of positive numbers
• Range: $(-\infty, \infty)$