how do you solve y=log2 3^x?

Question

I want to know how to do it using the substitution/table method where you substitue in numbers for one of the variables. Please show your process.

Answer 1

Remember that loga b = logb/loga so
y = log2 3^x = log3^x/log2 = log3^x/.301 and
.301y = log3^x

Now, if you know y calculate .301 times y and take 10 to that power.

10^(.301y) = 10^(log3x) = 3x and
10^(.301y)/3 = x

HTH

Doug

Answer 2

y = log(2)(3^x)
y = xlog(2)3
y = x * (log(3)/log(2))

Answer 3

i suggest a calculator