# Logarithmic equation with different bases

• Sep 25th 2013, 01:32 AM
corbomite1
Logarithmic equation with different bases
Would appreciate some guidance on the problem below:

Solve Log[base 25](6x + 25) + Log[base 5](x + 25) = 5, x a Natural number

To solve I tried converting to a single base using Log[base A]x = Log[base B]x / Log[base B]A which gives:

Log[base 25](6x + 25) = Log[base 5](6x + 25) / Log[base 5]25 = Log[base 5](6x+25) / 2

So the equation becomes:

Log[base 5](6x + 25) / 2 + Log[base 5](x + 25) = 5

Log[base 5](6x + 25) + 2Log[base 5](x + 25) = 10

Log[base 5](6x + 25) + Log[base 5](x + 25)^2 = 10

Log[base 5](6x + 25)(x + 25)^2 = 10

5^10 = (6x +25)(x +25)^2

This seems like a nasty polynomial to solve - is there a simpler approach to the problem?

Corbomite1
• Sep 25th 2013, 02:05 AM
Prove It
Re: Logarithmic equation with different bases
No that is exactly what you have to do.
• Sep 25th 2013, 02:28 AM
corbomite1
Re: Logarithmic equation with different bases
OK, thanks!
• Sep 25th 2013, 02:51 AM
Plato
Re: Logarithmic equation with different bases
Quote:

Originally Posted by corbomite1
Would appreciate some guidance on the problem below:
Solve Log[base 25](6x + 25) + Log[base 5](x + 25) = 5, x a Natural number

Here is some advice on notation.
You can learn to post symbols. It really is so easy. And it makes most of us more willing to find out how to help.
This subforum will help you with the code. Once you begin, you quickly learn the code.

[TEX]\log_{25}(6x+25)+\log_{5}(x+25)=5 [/TEX] gives \$\displaystyle \log_{25}(6x+25)+\log_5(x+25)=5 \$
If you click on the “go advanced tab” you should see \$\displaystyle \boxed{\Sigma} \$ on the tool-bar. That gives the [TEX]..[/TEX] wrap. Your LaTeX code goes between them.
• Sep 25th 2013, 02:56 AM
corbomite1
Re: Logarithmic equation with different bases
Understood, will do, thanks!

So the problem would read \$\displaystyle \log_{25}{(6x+25)} + log_5{(x+25)}=5\$ (Clapping)