Question 1036038: 2^3x+1=3^x-2
I cant figure out how to exactly set up this problem, so you csn help me out I woul really appreciate it.
So far I have 3log2+log2=xlog3-2log3
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! if your equation is 2^(3x+1) = 3^(x-2), then the solution is as follows.
take the log of both sides of the equation to get:
log(2^(3x+1)) = log(3^(x-2))
since log(a^b) = b*log(a), your equation becomes:
(3x+1) * log(2) = (x-2) * log(3)
simplify this equation to get:
3x * log(2) + log(2) = x * log(3) - 2 * log(3)
subtract x * log(3) from both sides of this equation and subtract log(2) from both sides of this equation to get:
3x * log(2) - x * log(3) = - 2 * log(3) - log(2)
factor out the x on the left side of this equation to get:
x * (3 * log(2) - log(3)) = -2 * log(3) - log(2)
divide both sides of this equation by (3 * log(2) - log(3)) to get:
x = (-2 * log(3) - log(2)) / (3 * log(2) - log(3)).
solve for x to get x = -2.946865368.
replace x in the original equation with -2.946865368 to confirm the solution is good.
the original equation is 2^(3x+1) = 3^(x-2).
after replacing x with -2.946865368, i got .0043625999 = .0043625999.
this confirmed the solution is correct.
you should use parentheses to make sure the person reading the problem knows what you are actually dealing with.
2^(3x+1) is a different expression than (2^3x) + 1.
without parentheses, 2^3x + 1 would be read as (2^3x) + 1.
i assumed you meant 2^(3x+1).
using the parentheses would have eliminated the need for me to make an assumption about what you are really trying to solve.
if, in fact, you were really trying to solve (2^3x) + 1 = (3^x) - 2, then let me know and i'll look at it again.
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