SOLUTION: Solve for x:{{{ 2(3^x) = 7(5^x) }}}

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Question 614548: Solve for x:+2%283%5Ex%29+=+7%285%5Ex%29+
Found 3 solutions by stanbon, sophxmai, lwsshak3:
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
2(3^x) = 7(5^x)
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(3^x)/(5^x) = 7/2
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(3/5)^x = 7/2
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x = log(7/2)/log(3/5)
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x = -2.4524..
Cheers,
Stan H.
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Answer by sophxmai(62) About Me  (Show Source):
You can put this solution on YOUR website!
take the log of both sides and then separate the 2(3^x) by the log(m)+log(n)=log(mn) rule.

+2%283%5Ex%29+=+7%285%5Ex%29+
log(2(3^x))=log(7(5^x))
log(2)+log(3^x)=log(7)+log(5^x)
log(2)+xlog(3)=log(7)+xlog(5)
xlog(3)-xlog(5)=log(7)-log(2)
x(log(3)-log(5))=log(7)-log(2)
x=(log(7)-log(2))/(log(3)-log(5))
x=-2.452427 <--i used a calculator

x=(log(7)-log(2))/(log(3)-log(5)) can be rewritten using the rule log(m)-log(n)=log(m/n)
x=log(7/2)/log(3/5)

Answer by lwsshak3(11628) About Me  (Show Source):
You can put this solution on YOUR website!
2(3^x) = 7(5^x)
2/7=5^x/3^x
2/7=(5/3)^x
log(2/7)=xlog(5/3)
x=log(2/7)/log(5/3)
x≈-2.452