SOLUTION: please help me solve this using logarithms 8^(3x+4) = 3^x stuff to the right of the ^'s are exponents....thanks!

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Question 35645: please help me solve this using logarithms
8^(3x+4) = 3^x
stuff to the right of the ^'s are exponents....thanks!
Answer by rapaljer(4671) About Me  (Show Source):
You can put this solution on YOUR website!
8%5E%283x%2B4%29+=+3%5Ex

First, take the ln of each side of the equation:
+ln++8%5E%283x%2B4%29+=+ln++3%5Ex

By the third law of exponents, "bring the exponents down" to become coefficients:
%283x%2B4%29+%2A+ln+8+=+x%2A+ln+3

Distributive property:
3x+%2A+ln+8+%2B+4+%2A+ln+8+=+x+%2A+ln+3

Get all the x terms on one side and the non-x terms on the other side. You might want to subtract +3x+ln+8 from each side:
+4%2A+ln+8+=+x+%2A+ln+3+-+3x%2A+ln+8

Factor out the x on the left side in order to get the x in one place, so you can solve for x.
4%2A+ln+8+=+x%28ln+3+-+3+%2A+ln+8%29+

Divide both sides by +%28ln+3+-+3+%2A+ln+8%29+

x+=+%284%2A+ln+8%29%2F%28+ln3+-+3+%2A+ln+8%29

Calculate with a calculator: x = -1.618 approximately.

Thank you so much for this problem!! I could put it on my final exam for College Algebra next week!!

R^2 at SCC