solve for x: 3^2x+3=7^x+1

Question

I did 2x+3=log0f3(7)+ logof3(7)
(2-log of 3(7))x=log of 3(7)-3
x=log of 3 (7)+log of 3 (3^-3)/log 0f 3 (3^2)-log of 3 (7)
x=log of 3 (7-1/3^3)/log of 3(9/7)
log of 3(7/27)/log of 3(9/7)
answer log of 9/7(7/27)
Is this correct? thank you for checking my work.

Answer 1

I assume you meant
3^(2x+3) = 7^(x+1)
Since the bases are in no way related, just take logs, in whatever base you like
(2x+3) log3 = (x+1) log7
2x log3 + 3log3 = x log7 + log7
(2log3 - log7)x = log7 - 3log3
x = (log7 - 3log3)/(2log3 - log7)
or, if you don't mind fractions,
x = log(7/27) / log(9/7)
so, you were right, though complicated, until your final step. In the expression "log x" the "log" part is not a factor that can cancel. If you want a "simpler" form, remember the base change formula, and you actually have

log_9/7 (7/27)

Answer 2

Thank you