Logarithm problem

**SVXX** · May 11th 2008, 10:11 PM

This equation has me at loggerheads...

$6(\log_{x}2 - \log_{4}x) + 7 = 0$

Even a hint would be appreciated!

**Moo** · May 11th 2008, 10:25 PM

Hello,

Originally Posted by SVXX

This equation has me at loggerheads...

$6(\log_{x}2 - \log_{4}x) + 7 = 0$

Even a hint would be appreciated!

Note that $\log_n x=\frac{\ln x}{\ln n}$

So $\log_x 2=\frac{\ln 2}{\ln x}$

And $\log_4 x=\frac{\ln x}{\ln 4}$

Multiply by ln(x), and ln(4) if you want to, then susbtitute y=ln x

**SVXX** · May 11th 2008, 10:35 PM

We haven't done natural logarithms in class yet, LOL.
Could I get some further tips for this? Or is there any other way to solve it?

**Moo** · May 11th 2008, 10:40 PM

Ok...

There is a formula :

$\log_b x=\frac{1}{\log_a b} \cdot \log_a x$ (took it from Wikipedia)

Taking $b=x$ and $a=4$ , you get :

$\log_x 2=\frac{1}{\log_4 x} \cdot \log_4 2$

So now you're working on base 4

This time, multiply by $\log_4 x$

**SVXX** · May 11th 2008, 11:08 PM

If I take $\log_{4}x$ as y, it will turn out to be a quadratic equation. So I guess we'll have two answers...

$ \log_{x}2 - \log_{4}x = -7/6$
$=> \log_{4}2/\_log_{4}x - \log_{4}x = -7/6$
$=> 1/2/y - y = -7/6$
$=> 1/2y - y = -7/6$
$=> 6y*y - 7y - 3 = 0$

The roots for y are 1.5 and 0.3333.
Taking y = 1.5,
$ \log_{4}x = 1.5 $
4 raised to 3/2 or 1.5 is x, or x = 8.
Merci beaucoup, Madame :P

**Moo** · May 11th 2008, 11:12 PM

Originally Posted by SVXX

If I take $\log_{4}x$ as y, it will turn out to be a quadratic equation. So I guess we'll have two answers...

$ \log_{x}2 - \log_{4}x = -7/6$
$=> \log_{4}2/\_log_{4}x - \log_{4}x = -7/6$
$=> 1/2/y - y = -7/6$
$=> 1/2y - y = -7/6$
$=> 6y*y - 7y - 3 = 0$

The roots for y are 1.5 and 0.3333.
Taking y = 1.5,
$ \log_{4}x = 1.5 $
4 raised to 3/2 or 1.5 is x, or x = 8.
Merci beaucoup, Madame :P

De rien

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