X variable in exponent, solve for x

**mwok** · December 17th 2008, 08:00 AM

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

Is it x = 2x - 1?

**Chop Suey** · December 17th 2008, 08:06 AM

Originally Posted by mwok

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

Is it x = 2x - 1?

NO! x is not equal to 2x - 1. Exponents are only equal to each other if the bases are the same.

Remember that $16 = 2^4$ and $8 = 2^3$

**TKHunny** · December 17th 2008, 08:08 AM

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.

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

Use those hints and show us what you get.

**mwok** · December 17th 2008, 08:12 AM

Ahh, thanks!

**mwok** · December 17th 2008, 08:22 AM

How would I do this one since it can't be made into a common base?

**Chop Suey** · December 17th 2008, 08:26 AM

Originally Posted by mwok

How would I do this one since it can't be made into a common base?

In this case, you have to know that $e^{\ln{(x)}} = x$ or $\ln{(e^x)} = x$ because $\ln{(x)}$ and $e^x$ are inverses.

**mwok** · December 17th 2008, 08:33 AM

Originally Posted by Chop Suey

In this case, you have to know that $e^{\ln{(x)}} = x$ or $\ln{(e^x)} = x$ because $\ln{(x)}$ and $e^x$ are inverses.

Still don't understand. Can you solve the problem so I can see how it works?

**Isomorphism** · December 17th 2008, 08:44 AM

Originally Posted by mwok

How would I do this one since it can't be made into a common base?

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:

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

Similarly:

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

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.

**mwok** · December 17th 2008, 08:46 AM

Ah, Thanks!

**Isomorphism** · December 17th 2008, 08:51 AM

Originally Posted by mwok

Ah, Thanks!

You are welcome... however do not ignore what Chop Suey said...

You could have got the answer in one step! Did you observe that $e^{\ln x} = x$ ?

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

**mwok** · December 17th 2008, 08:55 AM

Hmm, my solution is different from yours.

What happened to the ln(e)?

**Isomorphism** · December 17th 2008, 09:00 AM

Originally Posted by mwok

Hmm, my solution is different from yours.

What happened to the ln(e)?

Well ln stands for natural logarithm. Thus by definition, ln e = 1.

**mwok** · December 17th 2008, 09:02 AM

Originally Posted by Isomorphism

Well ln stands for natural algorithm. Thus by definition, ln e = 1.

Okay but how did you get rid of the ln(x+2) and ln(6) so it becomes x+2=6? Did you just cancel ln from both sides?

**Isomorphism** · December 17th 2008, 09:07 AM

Originally Posted by mwok

Okay but how did you get rid of the ln(x+2) and ln(6) so it becomes x+2=6? Did you just cancel ln from both sides?

Oops I actually used what Chop Suey pointed out

. I used the fact that ln has a unique inverse and thus $\ln x = \ln y \Rightarrow x = y$

On second thought, use what Chop Suey suggested.... Use $e^{\ln x} = x$

**mwok** · December 17th 2008, 09:11 AM

Originally Posted by Isomorphism

Oops I actually used what Chop Suey pointed out

. I used the fact that ln has a unique inverse and thus $\ln x = \ln y \Rightarrow x = y$

On second thought, use what Chop Suey suggested.... Use $e^{\ln x} = x$

Can you solve it using that method so I can see? I don't even know where the 2 and 6 goes.

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