What Are the Solutions to These Exponential Equations?

[chowder]

This is not my day.

Solve for x.

1.) 9^2x+1 = 27

2.) 10^5x = 1000


Thanks again!

 

[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

 

[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...

 

[VietDao29]

This is not my day.

Solve for x.

1.) 9^2x+1 = 27

2.) 10^5x = 1000


Thanks again!

You should use logarithm to solve the problems above.
Say if you need to solve a^x = b for x, where a and b are 2 constants. Then you can take log_a both sides, like this:
a^x = b
<=> log_a(a^x) = log_ab
<=> x = log_ab
I'll give you an example:
5^2x = 25
You can do this in 2 ways:
First way:
5^2x = 25 = 5²
<=> 2x = 2
<=> x = 1
Second way:
5^2x = 25
<=> log₅(5^2x) = log₅(25)
<=> 2x = 2
<=> x = 1
Ok, can you do the problems? Just try it again and see. :)

 

[J77]

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

 

[VietDao29]

Whoops, I am sorry, didn't pay attention when posting it.
Please don't take it personally.