1. ## exponential and logarithmic

I know how to solve these problems with the same base but i don't know what to do when the base isn't the same. Any help?

log x + log(x -
3) = 1

thanks!

2. Originally Posted by lizzy326
I know how to solve these problems with the same base but i don't know what to do when the base isn't the same. Any help?

log x + log(x -
3) = 1

thanks!

First apply the property $\displaystyle \ln a+\ln b= \ln(ab)$

We see the left side of the equation becomes $\displaystyle \log(x^2-3x)=1$

But note that $\displaystyle \log(10)=1$ [I'm assuming that when you refer to log, you mean it is of base 10]

So we see that $\displaystyle \log(x^2-3x)=\log(10)$

This is true only when $\displaystyle x^2-3x=10\implies x^2-3x-10=0$

Solve this quadratic equation and see what x values you get.

Does this make sense?

--Chris

3. Originally Posted by lizzy326
I know how to solve these problems with the same base but i don't know what to do when the base isn't the same. Any help?

log x + log(x -
3) = 1

thanks!

What do you mean "when the bases aren't the same"? In this example, you have the same base.

If you mean "$\displaystyle log_a(x)- log_b(x-3)$ then you have to change logarithm to the same base as the other.

$\displaystyle log_a(x)= \frac{log_b(x)}{log_b(a)}$ so
$\displaystyle log_a(x)- log_b(x-3)= \frac{log_b(x)}{log_b(a)}- log_b(x-3)= \frac{(log_b(x)- log_b(a)log_b(x-1)}{log_b(a)}$
$\displaystyle = \frac{log_b(x)- log_b((x-1)^{log_b(a)}}{log_b(a)= \frac{log_b(\frac{x}{(x-1)^{log_b(a)})}{log_b(a)}$

Not very pretty but that's the best you can do.