Question

solve the equation.\\(\log_{8}(3x + 1) + 16 = 18\\)\\(x = {?}\\)

Explanation:

Step1: Isolate the logarithmic term

Subtract 16 from both sides of the equation $\log_{8}(3x + 1)+16 = 18$ to get $\log_{8}(3x + 1)=18 - 16$.
Simplifying the right side gives $\log_{8}(3x + 1)=2$.

Step2: Convert logarithmic to exponential form

Recall that if $\log_{a}b = c$, then $b=a^{c}$. Applying this to $\log_{8}(3x + 1)=2$, we have $3x + 1 = 8^{2}$.

Step3: Solve for x

First, calculate $8^{2}=64$. So the equation becomes $3x + 1 = 64$.
Subtract 1 from both sides: $3x=64 - 1=63$.
Then divide both sides by 3: $x=\frac{63}{3}=21$.

Answer:

21