How would I solve for x?

9^x-1=27

Question

How would I solve for x? 

9^x-1=27

anonymous · Answer

$$9^{x-1} = 27$$

luigi0210 · Answer

Get common bases:
$$3^{2(x-1)}=3^3$$
$$2x-2=3$$
Finish it

anonymous · Answer

Oh okay thank you!

luigi0210 · Answer

Yup!

anonymous · Answer

I would use that same method with 3^2x+2= 8 right? @Luigi0210

luigi0210 · Answer

You might have to use logs, or just graph them.

anonymous · Answer

How would I use logs on this?

luigi0210 · Answer

$$3^{2x}=6$$
equals:
$$\log_{3} 6=2x$$

anonymous · Answer

where did the 6 come from

anonymous · Answer

i thought it would be 4

luigi0210 · Answer

on the 3^2x+2=8 right?
just subtract 2

anonymous · Answer

$$3^{2x+2}=8$$ i thought you would divide it but i guess i'm not sure

luigi0210 · Answer

Oh the numbers are all exponents? You should of mentioned that in the beginning :P
That as a log would look like this:
$$\log_{3} 8=2x+2$$

anonymous · Answer

oh okay thanks and then how would i take the log of that? I'm new to logs so I'm really not sure what I'm doing

luigi0210 · Answer

Then just subtract 2 and divide, make sure to not convert the log just yet because accuracy counts:
$$\frac{\log_{3} 8-2}{2}=x$$

anonymous · Answer

okay

anonymous · Answer

log38-1=x?

anonymous · Answer

$$\log _{3}8-1=x$$

luigi0210 · Answer

No, you have to do it all in one calculation..

anonymous · Answer

how

luigi0210 · Answer

You know, it might easier to graph this one for you ^_^'
The points of intersection are where your answer is at

anonymous · Answer

thank you ! if i plugged in the equation in my calculator would it give me the same results?

luigi0210 · Answer

It should