# exponents and logs

• May 14th 2007, 04:48 PM
turillian@gmail.com
exponents and logs
This type of question is kiiiiiling me right now.
Solve for x.
2^(1-x) = 3
Solve for x.
3^(2x-1) = 5

EDIT: these two are similar ones I can't find a solid approach to.
log base 2 of (log base 3 of (x)) = 4
and
10^5^x = 3

I thought I would brush up on my exponents and logs before I did the exponents and logs chapter in my calculus text, but I can't remember how to solve this kind of thing, thank you!
• May 14th 2007, 04:56 PM
qbkr21
Re;
Re:
• May 14th 2007, 05:01 PM
qbkr21
Re:
Re:
• May 14th 2007, 05:07 PM
alinailiescu
log_2(log_3(x))=4
log_3(x)=2^4
x=3^(2^4)
• May 14th 2007, 05:09 PM
alinailiescu
10^5^x=3
log_10(10^5^x)=log_10(3)
5^x=log_10(3)
log_5(5^x)=log_5(log_10(3))
x=log_5(log_10(3))
• May 15th 2007, 02:46 AM
topsquark
Actually there are two possible meanings to 10^5^x:
(10^5)^x
and
10^(5^x)

They are NOT equal so you need to use parenthesis!!

-Dan

Edit: I suppose by order of operations 10^5^x = (10^5)^x, but I would prefer the parenthesis for clarity.
• May 15th 2007, 05:05 AM
Soroban
Hello, turillian!

We can solve these without the base-change formula . . .

Quote:

Solve for x: .3^(2x-1) .= .5
Take logs of both sides. .The base doesn't matter; use ln.

. . . . . ln(3^{2x-1}) .= .ln(5)

Then: .(2x - 1)·ln(3) .= .ln(5)

. . . . . . . . . .2x - 1 .= .ln(5)/ln(3)

. . . . . . . . . . . . 2x .= .ln(5)/ln(3) + 1

. . . . . . . . . . . . .x .= .˝[ln(5)/ln(3) + 1]

Therefore: .x . .1.23248676

Quote:

log2[log3(x)] .= .4
Rewrite in exponential form.

We have: .log
3(x) .= .2^4 . . . . log3(x) .= .16

Rewrite again: .x .= .3^{16}

Therefore: .x .= .43,046,721

Quote:

10^5^x .= .3
Conventionally, a "stack" of exponents is read from the top.

We have: .10^(5^x) .= .3

Rewrite in logarithmic form: .5^x .= .log
10(3)

Rewrite again: .x .= .log
5[log10(3)]

Therefore: .x . .-0.459778294

• May 15th 2007, 01:52 PM
turillian@gmail.com
Thanks for the help everyone.
Yes, the stack of exponents was meant to be read from the top. It was
10^(5^x).