Please answer this one! How do i solve 2=3^(0.6934t). ?

Question

I think it has something to do with logs but i do not remember, please share knowledge!

Answer 1

take the log of both sides

log2=log(3^.6934t)

use a property of logs

log2=.6934tlog3

t=log2/(.6934log3)

plug that into a calculator

btw that log can be any base. ln, (e), log, (10), or log_n where n is any other non-0 non-1 positive integer

Answer 2

Remember:

log a^n = n log a

log ab = log a+log b

log a/b = log a-log b

Answer 3

log 2 = 0.6934 t (log 3)

0.6934 t = log 2 / log 3

t = [ log 2 / log 3 ] / 0.6934

t = 0.910---------(any log base may be used)

Answer 4

Use logarithms:

2 = 3^(0.6934t)

log 2 = log 3^(0.6934t)

Logarithmic laws say you can bring out the 0.6934t in front of the log 3:

log 2 = 0.6934t * log 3

Divide by log 3:

log 2 / log 3 = 0.6934t

Divide by 0.6934:

(log 2 / log 3) / 0.6934 = t

t ≈ 0.9099