how do u solve 2^x=3?

log2^x=log3
Using log laws:
xlog2=log3
x=log3/log2

just so you're not totally confused. both of the first two answers are correct.

you can take log [base anything] to solve.

so the first person use log (which is really log base 10) and the second person use ln (which is log base "e"). but anything would work.

for example, you can use log [base 2] if you wanted to:

log [base 2] 2^x = log [base 2] 3

x * log [base 2] 2 = log [base 2] 3 {you can move the exponent in front when dealing with logs}

now just divide by log [base 2] 2 to get:

x = (log [base 2] 3) / (log [base 2] 3)

x log 2 = log 3
x = log 3 / log 2
x = 1.58----------------any log base may be used

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

2^x = 3, so ln 2^x = ln 3
xln2 = ln 3
x=ln 2/ln 3