# Solve $11 \cdot 16^{1/(n-1)} = 16^{n/(n-1)} - 10$

This is probably an easy task for the users here, but I could not solve it.

$$11 \cdot 16^{1/(n-1)} = 16^{n/(n-1)} - 10$$

Wolfram Alpha gives the result $n= 5$.

What are the steps to solve this?

-

$11\times 16^{\frac{1}{n-1}}=16^{1+\frac{1}{n-1}}-10=16\times 16^{\frac{1}{n-1}}-10$, so $5\times 16^{\frac{1}{n-1}}=10$, so $16^{\frac{1}{n-1}}=2$, so $\frac{1}{n-1}=\frac{1}{4}$, so $n=5$.

-
Thank you for helping me. I will mark your answer as accepted as soon as possible. –  Tyymo Dec 12 '12 at 12:58

$$11 \cdot 16^{1/(n-1)} = 16^{n/(n-1)} - 10$$ $$11 \cdot 16^{1/(n-1)} = 16^{1+1/(n-1)} - 10$$

$$11 \cdot 16^{1/(n-1)} = 16\cdot16^{1/(n-1)} - 10$$

$$16\cdot 16^{1/(n-1)} - 11\cdot16^{1/(n-1)}=10$$

$$5\cdot 16^{1/(n-1)}=10$$

$$16^{1/(n-1)}=2=16^{1/4}$$ $$1/(n-1)=1/4$$ $$n=5$$

-