Logarithmic equations

**Furyan** · November 18th 2012, 09:13 AM

Hello

The question is, solve:

$2\log_{x}4 - \log_{2}x = 3$

Changing the base I get:

$\log_{x}16 - \dfrac{\log_{x}x}{\log_{x}2} = 3$

$\log_{x}16 - \dfrac{1}{\log_{x}2} = 3$

I don't know how to proceed from here.

I know that the difference between two logs with the same base can be written as a division, but the reciprocal is throwing me off.

I would greatly appreciate some help.

Thank you.

**skeeter** · November 18th 2012, 09:20 AM

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

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

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

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

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

$4 - u^2 = 3u$

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

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

**Furyan** · November 18th 2012, 09:49 AM

Hello skeeter,

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

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

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

Brilliant! Thank you very much.

Thread: Logarithmic equations

LinkBack

Thread Tools

Display

Logarithmic equations

Re: Logarithmic equations

Re: Logarithmic equations

Similar Math Help Forum Discussions

Logarithmic Equations

Logarithmic Equations

Logarithmic equations

Help with Logarithmic Equations

logarithmic equations

Search Tags