I need help with logarithms.

log_(2x+6) 8 = 3
and
log_27 9

Question

I need help with logarithms.

log_(2x+6) 8 = 3
and 
log_27 9

anonymous · Answer

So for the first one using log_a (b) = c to exponential form. a^c = b. (2x+6)^3=8. (2x+6)(2x+6)(2x+6). Luckily the other one is much simpler. Again rewritting it as 27^x=9. First what I'll do is I recognize 27 as 3^3. That would still give me 3^3x =9. But now I also happen to know that 9 is 3^2. At which point 3^3x = 3^2. At which point you can solve 3x=2.

anonymous · Answer

I found the second one is 3/2 or 1.5, right?

anonymous · Answer

And for the first one does that mean I find (2x+6)(2x+6)(2x+6) = 8?

anonymous · Answer

I got -2 for the first answer when I solved it that way. Is that right?

loser66 · Answer

continue 11n stuff |dw:1387061772020:dw|