# Thread: 9^x = 81. Determine x.

1. ## 9^x = 81. Determine x.

$9^x = 81.$ Determine x.

Now I found that the answer is 2 via trial and error. However, I wanted to know whether or not there was a formal mathematical procedure which proves my answer. I haven't been taught this unit yet, for the unit I'm working on now (involving inverse functions) hasn't been assigned yet. I've just been working ahead.

$9^x = 81.$ Determine x.

Now I found that the answer is 2 via trial and error. However, I wanted to know whether or not there was a formal mathematical procedure which proves my answer. I haven't been taught this unit yet, for the unit I'm working on now (involving inverse functions) hasn't been assigned yet. I've just been working ahead.
1. 9^x = 81 = 9^2

If the bases are equal the exponents must be equal so $x = 2$

2. $x\, \log_9(9) = \log_9 (81)$. Hence $x=2$

3. "Trial and Error" is a perfecly good "mathematical procedure". What more do you want?

4. Thanks for the replies. I was just wondering if there were certain steps to be shown in order to get the answer. One of the questions asks:

$8^m = 256$, determine m.

The answer is $m = \frac {8}{3}$.

How would I determine the answer of this question, without using the back of the book?

Thanks for the replies. I was just wondering if there were certain steps to be shown in order to get the answer. One of the questions asks:

$8^m = 256$, determine m.

The answer is $m = \frac {8}{3}$.

How would I determine the answer of this question, without using the back of the book?
$8^m = 256$

${(2^3)^m = 2^8}$

$2^{3m} = 2^8$

therefore, by the equality of bases, you get

$3m = 8$

$m = \frac{8}{3}$

Is that clear enough?

Thanks for the replies. I was just wondering if there were certain steps to be shown in order to get the answer. One of the questions asks:

$8^m = 256$, determine m.

The answer is $m = \frac {8}{3}$.

How would I determine the answer of this question, without using the back of the book?
This probably isn't the most general method, but

$8^m=256=2^8$

$(2^3)^m=2^8$

$2^{3m}=2^8$

$3m=8$

$m=\frac{3}{8}$

Since I do a bit of computer programming, some powers of 2 are very easy to recognize... someone will probably post a more general method after me, but tricks like this can often be used (express the base, in this case 8, as a smaller base raised to a power, in this case $2^3$).

7. Yes, that's exactly what I'm looking for! Thx!

8. If the problem does not come out so evenly, like maybe 3^x=17, you can just take the log (to any base) of both sides and solve for x. So, if a and b are constants, and

$a^x=b$

the solution is:

$\log{a^x}=\log{b}$

$x\log{a}=\log{b}$

$x=\frac{\log{b}}{\log{a}}$

Of course, that's more work than you need to do for a case like 9^x=81.

- Hollywood