# A couple more

1. Sep 13, 2006

### chowder

This is not my day.

Solve for x.

1.) 9^2x+1 = 27

2.) 10^5x = 1000

Thanks again!

2. Sep 14, 2006

### steve23063

assuming base 2:
log30=4x is equivalent to writing 2^4x=30
log30 is equivalent to writing (log30)/(log2)
This means (log30)/(log2)=4x

assuming base 8:
log12=5x+4 is equivalent to writing 8^(5x+4)=12
log12 is equivalent to writing (log12)/(log8)
This means (log12)/(log8)=5x+4

I hope those hints are enough. If you still can't figure it out just ask

3. Sep 14, 2006

### J77

You can also think about equating exponents:

Just to clarify, but in your first question, you must mean 9^(2x+1)=27

As an example,

4^(3x-1)=8

is equiv. to

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

equating exponents:

6x-2=3

which implies:

x=5/4

Now do the same for your problems...

4. Sep 14, 2006

### VietDao29

You should use logarithm to solve the problems above.
Say if you need to solve ax = b for x, where a and b are 2 constants. Then you can take loga both sides, like this:
ax = b
<=> loga(ax) = logab
<=> x = logab
I'll give you an example:
52x = 25
You can do this in 2 ways:
First way:
52x = 25 = 52
<=> 2x = 2
<=> x = 1
Second way:
52x = 25
<=> log5(52x) = log5(25)
<=> 2x = 2
<=> x = 1
Ok, can you do the problems? Just try it again and see. :)

Last edited: Sep 14, 2006
5. Sep 14, 2006

### J77

Did you not just repeat the two methods which me and steve gave?

6. Sep 14, 2006

### VietDao29

Whoops, I am sorry, didn't pay attention when posting it.