Answer
a. $\quad3$
b. $\quad\dfrac{5}{2}$
c. $\quad-\dfrac{5}{6}$
Work Step by Step
(a) Use the definition of logarithm, which states that $\log_b {x} = y\longleftrightarrow b^{y} = x$, to write an exponential equation. In this exercise, the base $b$ is $5$, $y$ is the exponent, and $x$ is $125$:
Let $y=\log_5{125}$. Then,
$$5^{y} = 125$$
Rewrite terms so that they both have the same base:
$$5^{y} = 5^{3}$$
If two numbers having the same base are equal, that means that their exponents are also the same, so set the exponents equal to one another to solve for $y$:
$$y = 3$$
(b). Use the definition of logarithm, which states that , to write an exponential equation. In this exercise, the base $b$ is $4$, $y$ is the exponent, and $x$ is $32$:
Let $y=\log_4{32}$. Then,
$$4^{y} = 32$$
Rewrite terms so that they both have the same base:
$$(2^{2})^{y} = 2^{5}$$
When raising a power to a power, multiply the exponents, keeping the base as-is:
$$2^{2y} = 2^{5}$$
If two numbers having the same base are equal, that means that their exponents are also the same, so set the exponents equal to one another to solve for $y$:
$$2y = 5$$
Divide both sides by $2$ to solve for $y$:
$$y = \frac{5}{2}$$
(c) Use the definition of logarithm to write an exponential equation. In this exercise, the base $b$ is $64$, $y$ is the exponent, and $x$ is $\frac{1}{32}$:
Let $y=\log_{64}{\frac{1}{32}}$. Then,
$$64^{y} = \frac{1}{32}$$
Rewrite $\frac{1}{32}$ as an exponential expression:
$$64^{y} = 32^{-1}$$
Rewrite terms so that they both have the same base:
$$(2^{6})^{y} = (2^{5})^{-1}$$
When raising a power to a power, multiply the exponents, keeping the base as-is:
$$2^{6y} = 2^{-5}$$
If two numbers having the same base are equal, that means that their exponents are also the same, so set the exponents equal to one another to solve for $y$:
$$6y = -5$$
Divide both sides by $6$ to solve for $y$:
$$y = -\frac{5}{6}$$
