Re: Logarithmic equations

$\displaystyle 2\log_x{4} - \log_2{x} = 3$

$\displaystyle 2 \cdot \frac{\log_2{4}}{\log_2{x}} - \log_2{x} = 3$

$\displaystyle \frac{4}{\log_2{x}} - \log_2{x} = 3$

let $\displaystyle u = \log_2{x}$ ...

$\displaystyle \frac{4}{u} - u = 3$

$\displaystyle 4 - u^2 = 3u$

$\displaystyle 0 = u^2 + 3u - 4$

solve the quadratic for u , then back substitute to solve for x.

Re: Logarithmic equations

Hello skeeter,

I had also tried changing the base of the log to 2 and got:

$\displaystyle \frac{4}{\log_2{x}} - \log_2{x} = 3$

but I hadn't thought of using a substitution and then solving.

Brilliant! Thank you very much. (Bow)