Exponential and Logarithmic Equations

• May 18th 2013, 02:48 PM
Cake
Exponential and Logarithmic Equations
Hey guys,

log(x+3) - logx = 1

so I got 1 = (x+3) / (x)

and the book says the answer is 1/3

I don't know how it got this :(

any help? Thanks
• May 18th 2013, 02:55 PM
HallsofIvy
Re: Exponential and Logarithmic Equations
Is that the common logarithm? Base 10? If so then you have $log(x+3)- log(x)= log\left(\frac{x+ 3}{x}\right)= 1$ and then "get rid of" the logarithm by taking 10 to the power of each side. Instead of getting $\frac{x+ 3}{x}= 1$ you should have $\frac{x+ 3}{x}= 10^1= 10$.

Can you solve $\frac{x+ 3}{x}= 10$?
• May 18th 2013, 03:01 PM
Plato
Re: Exponential and Logarithmic Equations
Quote:

Originally Posted by Cake
log(x+3) - logx = 1
and the book says the answer is 1/3 I don't know how it got this

.

The answer depends upon the base.

If ${\log _b}(x + 3) - {\log _b}(x) = 1$ then ${\log _b}\left( {\frac{{x + 3}}{x}} \right) = 1$.

Then $\frac{x+3}{x}=b$.

So I don't know where that answer comes from.
• May 18th 2013, 03:06 PM
Cake
Re: Exponential and Logarithmic Equations
Correct, it is base 10. Ahh I see. So whenever I have " = 1 " and with a base of 10. It automatically equals to 10^1?

and yes, thank you. That gives me 1 / 3. Thanks a lot for clearing this up. I would have never gotten this.