Question

I need help with this math question

Answer 1

Answer:

B. 5

Step-by-step explanation:

Step 1: Rewrite the equation a bit

$\frac{7^\frac{3}{4}}{7^\frac{x}{8}}=\sqrt[8]{7}\\7^\frac{3}{4}^-^\frac{x}{8}=7^\frac{1}{8}$

Step 2: Place a logarithm base 7 on both sides

$7^\frac{3}{4}^-^\frac{x}{8}=7^\frac{1}{8}\\log_77^\frac{3}{4}^-^\frac{x}{8}=log_77^\frac{1}{8}\\(\frac{3}{4}-\frac{x}{8})log_77=\frac{1}{8}log_77\\(\frac{3}{4}-\frac{x}{8})1=\frac{1}{8}1\\\frac{3}{4}-\frac{x}{8}=\frac{1}{8}$

Step 3: Solve for x

$\frac{3}{4}-\frac{x}{8}=\frac{1}{8}\\\frac{6}{8}-\frac{x}{8}=\frac{1}{8}\\-\frac{x}{8}=-\frac{5}{8}\\\frac{x}{8}=\frac{5}{8}\\x=5$

Neat exponent question :)