Re:
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!
Hello, turillian!
We can solve these without the base-change formula . . .
Take logs of both sides. .The base doesn't matter; use ln.Solve for x: .3^(2x-1) .= .5
. . . . . 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
Rewrite in exponential form.log2[log3(x)] .= .4
We have: .log3(x) .= .2^4 . . → . . log3(x) .= .16
Rewrite again: .x .= .3^{16}
Therefore: .x .= .43,046,721
Conventionally, a "stack" of exponents is read from the top.10^5^x .= .3
We have: .10^(5^x) .= .3
Rewrite in logarithmic form: .5^x .= .log10(3)
Rewrite again: .x .= .log5[log10(3)]
Therefore: .x .≈ .-0.459778294