1. ## Basic Logarithm Problem

Alright, this one particular question is driving me nuts, I've tried it with many different approaches, but they keep producing undefined results. It's probably something simple that I'm overlooking, but if someone could help show me the proper approach to this, it would help a lot.

$9^(2x)=10^x$

**Edit** Sorry, the 2x is supposed to be a power of 9.

2. Originally Posted by unk2100
Alright, this one particular question is driving me nuts, I've tried it with many different approaches, but they keep producing undefined results. It's probably something simple that I'm overlooking, but if someone could help show me the proper approach to this, it would help a lot.

$9^(2x)=10^x$

**Edit** Sorry, the 2x is supposed to be a power of 9.
$9^{2x} = 10^x$

$81^x = 10^x$

$x = 0$

3. Originally Posted by unk2100
Alright, this one particular question is driving me nuts, I've tried it with many different approaches, but they keep producing undefined results. It's probably something simple that I'm overlooking, but if someone could help show me the proper approach to this, it would help a lot.

$9^(2x)=10^x$

**Edit** Sorry, the 2x is supposed to be a power of 9.
$9^{2x} = (3^2)^{2x} = 3^{4x}$

$3^{4x}=10^x$

Take logs

$4x\, log_{10}(3) = x$

$x(4\,log_{10}3 - 1) = 0$

$x = 0$

4. wow, you guys are quick. Thanks.