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Mathematics
OpenStudy (anonymous):

Solve 5^(3x) = 2^(x^2-1)

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!

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?

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

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

OpenStudy (anonymous):

I'm asking other people to take a look at this lol

OpenStudy (anonymous):

answer = 0.14 and 6.81 LOL

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