# Thread: solving exponents with logs

1. ## solving exponents with logs

solve:
$2^{2x} = 5^x$

i went this far:
$\log{2^{2x}} = \log{5^x}$

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

how would you finish it?

2. You don't even need logs, just note that

$2^{2x} = 5^x \implies \left( {\frac{4}
{5}} \right)^x = 1.$

Now, what's the next step?

3. $anything^0 = 1$

but wouldnt you get a different answer with logs?

4. Originally Posted by Krizalid
$2^{2x} = 5^x \implies \left( {\frac{4}
{5}} \right)^x = 1.$
doesnt this just turn every problem of its kind into x = 0?

for instance with random numbers:

$12345^{67x} = 5432^x \implies \left( {\frac{12345^{67}}
{5432}} \right)^x = 1 \implies x = 0$

5. Originally Posted by incyt3s
$anything^0 = 1$

but wouldnt you get a different answer with logs?
no you would not. Krizalid is right, logs are definitely not needed. however, if you do want to force the issue...

we have $4^x = 5^x$

take the log of both sides

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

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

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

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

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

this second option is an impossibility, thus $x = 0$ as we got from Krizalid's much simpler solution

6. Originally Posted by incyt3s
doesnt this just turn every problem of its kind into x = 0?

for instance with random numbers:

$12345^{67x} = 5432^x \implies \left( {\frac{12345^{67}}
{5432}} \right)^x = 1 \implies x = 0$
technically no, but i guess in general yes.

here's one where x is not zero (well, at least that's not the only solution)

$4^{3x} = 64^x$

7. 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 $4^{3x} = 64^x$, we can see we cant use the above idea.