Find x in terms of base 10 logs

Question

Step by step, how do you answer this question:

Find in terms of logarithms to the base 10, the exact value of x such that 3^2x+1=2^x-4.

(Also, do you know any good sites/resources for A-Level maths (especially logs) revision?).

Answer 1

3^(2x+1)=2^(x-4)

log3^(2x+1)=log2^(x-4)

(2x+1)log3=(x-4)log2

(2x+1)/(x-4)=log2/log3

cross multiply

log2(x-4)=log3(2x+1)

distribute

xlog2-4log2=2xlog3+log3

xlog2-4log2=xlog3^2+log3

xlog2-4log2=xlog9+log3

xlog2-xlog9=4log2+log3

x(log2-log9)=4log2+log3

x=(4log2+log3)/(log2-log9) is the exact answer

if you want to use your calculator and find all of the logs, you will get x=-2.57380636

substituting to check the answer x=-2.57380636

3^(2*[-2.57380636]+1)=2^(-2.57380636-4)

3^(-4.14761272)=2^(-6.57380636)

1/(3^4.14761272)=1/(2^6.57380636)

1/95.26079137≈1/95.26050951