exponential and logarithmic

**lizzy326** · October 29th 2008, 07:34 AM

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!

**Chris L T521** · October 29th 2008, 07:47 AM

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 $\ln a+\ln b= \ln(ab)$

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

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

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

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

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

Does this make sense?

--Chris

**HallsofIvy** · October 29th 2008, 09:01 AM

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

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

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

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