Thread: 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!

Re:

Re:

log_2(log_3(x))=4
log_3(x)=2^4
x=3^(2^4)

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))

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.

Hello, turillian!

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

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 .= .½[ln(5)/ln(3) + 1]


Therefore: .x .≈ .1.23248676

log2[log3(x)] .= .4

Rewrite in exponential form.

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

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

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

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 .= .log10(3)

Rewrite again: .x .= .log5[log10(3)]

Therefore: .x .≈ .-0.459778294

Thanks for the help everyone.
Yes, the stack of exponents was meant to be read from the top. It was
10^(5^x).