solve: $log_{7}(2x - 1)=2$\na $64\frac{1}{2}$\nb $63\frac{1}{2}$\nc 25\nd 24

Explanation:

Step1: Convert to exponential form

By the definition of logarithms, if $\log_{a}b = c$, then $b=a^{c}$. Here $a = 8$, $b = 2x - 1$, $c = 2$. So $2x-1=8^{2}$.

Step2: Calculate $8^{2}$

$8^{2}=64$, so the equation becomes $2x - 1=64$.

Step3: Solve for $x$

Add 1 to both sides: $2x=64 + 1=65$. Then divide both sides by 2, $x=\frac{65}{2}=32\frac{1}{2}$. But there is a mistake above, it should be $\log_{7}(2x - 1)=2$. Re - do:

Step1: Convert to exponential form

If $\log_{7}(2x - 1)=2$, then $2x-1 = 7^{2}$ by the definition of logarithms ($\log_{a}b=c\Rightarrow b=a^{c}$).

Step2: Calculate $7^{2}$

$7^{2}=49$, so the equation is $2x-1=49$.

Step3: Solve for $x$

Add 1 to both sides: $2x=49 + 1=50$. Divide both sides by 2, $x = 25$.

Answer:

C. 25