Attached is the equation. Could someone show me the steps? I know you had to cross-multiply something.

Is it x = 2x - 1?

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- Dec 17th 2008, 08:00 AMmwokX variable in exponent, solve for x
Attached is the equation. Could someone show me the steps? I know you had to cross-multiply something.

Is it x = 2x - 1? - Dec 17th 2008, 08:06 AMChop Suey
- Dec 17th 2008, 08:08 AMTKHunny
Please get the concept of "cross multiply" out of your life. It is not your friend. The fact that it doesn't even remotely apply to this problem suggests to me that it is as confusing as I like to say it is.

You almost have this one, but failed to spot that the bases were different. If they were both 8 or both 16, you would be right on. You must change to a common base.

$\displaystyle 16 = 2^{4}$ and $\displaystyle 8 = 2^{3}$

Use those hints and show us what you get. - Dec 17th 2008, 08:12 AMmwok
Ahh, thanks!

- Dec 17th 2008, 08:22 AMmwok
How would I do this one since it can't be made into a common base?

- Dec 17th 2008, 08:26 AMChop Suey
- Dec 17th 2008, 08:33 AMmwok
- Dec 17th 2008, 08:44 AMIsomorphism
Its actually a problem that is very similar to the first one. Whenever you see a power on one side and some number on the other, always take logs on both sides. And then try to play with the logs.

For example:

$\displaystyle 16^x = 8^{2x-1} \Rightarrow x \log 16 = (2x - 1) \log 8 $$\displaystyle \Rightarrow x (4\log 2) = (2x - 1) (3\log 2) \Rightarrow 4x = 3(2x-1)$

Similarly:

$\displaystyle e^{\ln (x+2)} = 6 \Rightarrow \ln e^{\ln (x+2)} = \ln 6 \Rightarrow$$\displaystyle \ln (x+2) \ln e = \ln 6 \Rightarrow \ln (x+2) = \ln 6 \Rightarrow x+2 = 6$

This method is very mechanical. Identifying both sides as powers of the same base is faster, if you get used to it. But you will need practice before you can do that. - Dec 17th 2008, 08:46 AMmwok
Ah, Thanks!

- Dec 17th 2008, 08:51 AMIsomorphism
- Dec 17th 2008, 08:55 AMmwok
Hmm, my solution is different from yours.

What happened to the ln(e)? - Dec 17th 2008, 09:00 AMIsomorphism
- Dec 17th 2008, 09:02 AMmwok
- Dec 17th 2008, 09:07 AMIsomorphism
- Dec 17th 2008, 09:11 AMmwok