I'll kick it off for you.
now you need to change bases to either 4 or 2.
Just in case there were others who did not understand how to proceed from what pickslides had written, this is how one would finish the problem:
log_2 (x+1) = 1 + log_4 (3x-2)
4^log_2 (x+1) = 12x-8 (convert the above to exponential form)
2^log_2 (x+1)^2 = 12x-8 (convert the 4 into 2^2 and then use the property of logarithms to move the latter 2 (the one after the carrot) from front of the log sign so that it can square (x+1))
(x+1)^2 = 12x-8 (cancel out the other 2_log 2 based on the property of logarithms)
x^2 + 2x +1 = 12x-8 (solve the quadratic equation)
x^2 -10x + 9 = 0
x=1, x=9
Since neither solution is negative, both solutions are valid and will work. I have checked using a calculator to make sure.
I would have done the first problem is a slightly different way. From , subtract from both sides to get .
Now change one of the bases: means and from that . Thus, and the equation becomes .
And that gives . Now you can write that as , square both sides and finish as before.
As for which is, indeed, the same as , I would recommend that now you use the fact that if , then .
That is, you now have and that is the same as saying . Solve that for x by using whatever logarithm is handy, either "common" or "natural. so . That cannot be done "by hand". Use either common log or natural log, whichever is easier to do on your calculator.