solve:

$\displaystyle 2^{2x} = 5^x$

i went this far:

$\displaystyle \log{2^{2x}} = \log{5^x}$

$\displaystyle 2x\log{2} = x\log{5}$

how would you finish it?

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- Dec 16th 2007, 05:34 PMincyt3ssolving exponents with logs
solve:

$\displaystyle 2^{2x} = 5^x$

i went this far:

$\displaystyle \log{2^{2x}} = \log{5^x}$

$\displaystyle 2x\log{2} = x\log{5}$

how would you finish it? - Dec 16th 2007, 06:12 PMKrizalid
You don't even need logs, just note that

$\displaystyle 2^{2x} = 5^x \implies \left( {\frac{4}

{5}} \right)^x = 1.$

Now, what's the next step? - Dec 16th 2007, 09:48 PMincyt3s
$\displaystyle anything^0 = 1$

but wouldnt you get a different answer with logs? - Dec 16th 2007, 10:02 PMincyt3s
- Dec 16th 2007, 10:03 PMJhevon
no you would not. Krizalid is right, logs are definitely not needed. however, if you do want to force the issue...

we have $\displaystyle 4^x = 5^x$

take the log of both sides

$\displaystyle \Rightarrow \log 4^x = \log 5^x$

$\displaystyle \Rightarrow x \log 4 = x \log 5$

$\displaystyle \Rightarrow x \log 4 - x \log 5 = 0$

$\displaystyle \Rightarrow x ( \log 4 - \log 5) = 0$

$\displaystyle \Rightarrow x = 0 \mbox{ or } \log 4 - \log 5 = 0$

this second option is an impossibility, thus $\displaystyle x = 0$ as we got from Krizalid's much simpler solution - Dec 16th 2007, 10:05 PMJhevon
- Dec 17th 2007, 09:48 PMIsomorphism
Another way to see it is that a power of 2 can never have a factor of 5 or the other way.

Thus technically x=0 is the only solution.

But with $\displaystyle 4^{3x} = 64^x$, we can see we cant use the above idea.