How do you solve $7-4logx=10$ ?

Question

2235 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-12-19T14:00:09

$10^(-3/4)$

$7-4logx=10$

Subtract $7$ from both sides.

$-4logx=3$

Divide both sides by $-4$ .

$logx=-3/4$

Recall that $logx=log_10 x$ .

Exponentiate both sides to undo the logarithm

$10^(log_10x)=10^(-3/4)$

$x=10^(-3/4)$

This can be written in a variety of different ways.

$10^(-3/4)=1/(10^(3/4))=1/root4(10^3)=1/root4(1000)~=0.1778$

How do you solve 7-4logx=107-4logx=10?