Basic Logarithm Problem

**unk2100** · November 27th 2009, 01:16 PM

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.

**skeeter** · November 27th 2009, 01:21 PM

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.

$9^{2x} = 10^x$

$81^x = 10^x$

$x = 0$

**e^(i*pi)** · November 27th 2009, 01:23 PM

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.