log and exponential forms with finding x

**screamironic** · March 30th 2009, 04:41 PM

We have been working with logs and didn't get any examples on how to do these types of problems. I was just wondering if someone could help me out on one of them.

5^(x-1)=2^x

**skeeter** · March 30th 2009, 06:05 PM

Originally Posted by screamironic

We have been working with logs and didn't get any examples on how to do these types of problems. I was just wondering if someone could help me out on one of them.

5^(x-1)=2^x

$5^{x-1} = 2^x$

take the log of both sides ...

$\log(5^{x-1}) = \log(2^x)$

use the power rule for logs ...

$(x-1)\log(5) = x\log(2)$

distribute the left side ...

$x\log(5) - \log(5) = x\log(2)$

get terms with x on the same side ...

$x\log(5) - x\log(2) = \log(5)$

factor out the x ...

$x[\log(5) - \log(2)] = \log(5)$

solve for x ...

$x = \frac{\log(5)}{\log(5) - \log(2)} = \frac{\log(5)}{\log(2.5)}$

**stapel** · March 31st 2009, 08:45 AM

Originally Posted by screamironic

We have been working with logs and didn't get any examples on how to do these types of problems.

Ouch! So you're needing help with understanding the topic first....

To learn, in general, how to solve this sort of equation, try here. Once you've studied that, the worked solution in the previous reply should make perfect sense!

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