Question

determine the value of y for each of the following.
a) $\log_y 4 = 2$
value
b) $\log_8 y = \frac{1}{3}$
value
c) $\log_7 49 = y$
value

Explanation:

Part (a)

Step 1: Recall the logarithmic definition

The logarithmic equation $\log_{y} 4 = 2$ can be converted to its exponential form using the definition $\log_{a} b = c \implies a^{c}=b$. So, we have $y^{2}=4$.

Step 2: Solve for $y$

We solve the equation $y^{2} = 4$. Taking the square root of both sides, we get $y=\pm2$. But in the logarithm $\log_{y}4$, the base $y>0$ and $y
eq 1$. So we discard $y = - 2$. Thus, $y = 2$.

Part (b)

Step 1: Recall the logarithmic definition

The equation $\log_{8}y=\frac{1}{3}$ can be converted to exponential form. Using $\log_{a}b = c\implies a^{c}=b$, we have $8^{\frac{1}{3}}=y$.

Step 2: Evaluate $8^{\frac{1}{3}}$

We know that $8 = 2^{3}$, so $8^{\frac{1}{3}}=(2^{3})^{\frac{1}{3}}$. Using the power - of - a - power rule $(a^{m})^{n}=a^{mn}$, we get $(2^{3})^{\frac{1}{3}}=2^{3\times\frac{1}{3}}=2^{1}=2$. So $y = 2$.

Part (c)

Step 1: Recall the logarithmic definition

The equation $\log_{7}49=y$ can be converted to exponential form. Using $\log_{a}b = c\implies a^{c}=b$, we have $7^{y}=49$.

Step 2: Solve for $y$

We know that $49 = 7^{2}$. So the equation $7^{y}=7^{2}$ implies that $y = 2$ (since if $a^{m}=a^{n}$, then $m = n$ for $a>0,a
eq1$).

Answer:

a) The value of $y$ is $\boldsymbol{2}$.

b) The value of $y$ is $\boldsymbol{2}$.

c) The value of $y$ is $\boldsymbol{2}$.