Help with logs?

Question

2^5x-1=3^2x+1

All i need to see is how to do this problem so i can figure out how to do the rest >.<

Answer 1

Taking logs of both sides, (5x-1)log(2) = (2x+1)log(3), so x[5log(2)-2log(3)]=log(3)+log(2), so x[log(32/9)]=log(6). Solve for x.

Steve

EDIT - If you mean (2^5x)-1 = (3^2x)+1, then you have 32^x-9^x=2, which has one solution at x=0.443392.

Answer 2

I will assume that 5x-1 is all an exponent and so is 2x+1.

Take the log of both sides. That turns the exponent into a factor:

Log(2^5x-1) = Log (3^2x+1)

(5x-1) log 2 = (2x+1) log 3

distribute:

5x log 2 - log 2 = 2x log 3 + log 3

Move the terms with x to the left and the terms without x to right

Then distribute (factor) out the x, divide, and calculate

Answer 3

2^(5x-1) = 3^(2x+1)
log (2^(5x-1)) = log (3^(2x+1)) ............ (at base 2)
(5x-1)log 2 = (2x+1)log 3
(5x-1) = (2x+1)log 3
(5x-1)/(2x+1) = log 3

₢

Answer 4

2^5x-1=3^2x+1
2^5x=3^2x+2---add 1 to both sides
5xlog2=2xlog3+log2---take logs of both sides
5xlog2-2xlog3=log2---subtract 2xlog3 from both sides
x(5log2-2log3)=log2---factor out an x
x=log2/(5log2-2log3)---divide both sides by 5log2-2log3
x=0.5464---use calculator to calculate out; x is about 0.5464