Mathematics

OpenStudy (anonymous):
Solve
5^(3x) = 2^(x^2-1)

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OpenStudy (anonymous):
is there an easy way? LOL

OpenStudy (anonymous):
Is it this? \[5^{3x} = 2^{x ^{2}-1}\]

OpenStudy (anonymous):
YUP!

OpenStudy (anonymous):
Something is telling me to use logs...brb

OpenStudy (anonymous):
YUP!

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OpenStudy (anonymous):
i kinda know how to do it.. but i was wondering if theres an easier way hehe

OpenStudy (anonymous):
probably not

OpenStudy (anonymous):
\[3xLN5 = (X^2-1)LN2 \] i think

OpenStudy (anonymous):
probably not
\[(3x )\log (5) = (x ^{2}-1)\log(2)\]

OpenStudy (anonymous):
log and LN is not the same thing?

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OpenStudy (anonymous):
You can use either one and it won't affect the answer

OpenStudy (anonymous):
Wow you're right this one is a stubborn one

OpenStudy (anonymous):
i figure it out

OpenStudy (anonymous):
\[3xln5 = \ln(x^2)-\ln2\]

OpenStudy (anonymous):
it looks like a quadratic

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OpenStudy (anonymous):
Yep it is a quadratic equation, the type that nightmares are made of hahahaha

OpenStudy (anonymous):
omg I'm at this point now: \[\log(2)x^2 - 3(\log5)x - \log2 = 0\]

OpenStudy (anonymous):
omg I do not want to plug that into the quadratic formula...no wonder you asked if there was an easier way...

OpenStudy (anonymous):
haha YUP! thats what i got too

OpenStudy (anonymous):
i plugged it in and its horrible

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OpenStudy (anonymous):
I'm asking other people to take a look at this lol

OpenStudy (anonymous):
answer = 0.14 and 6.81 LOL