Exact Logarithmic Problems

its 6 you multiply the two powers

juicybebe101:

its 6 you multiply the two powers

Hello! The answer "6" is wrong. This could be 6 if it was e^((ln(3)+ln(2)).
You can write e^((ln(3)*ln(2)) as:

Dms:

juicybebe101:

its 6 you multiply the two powers

Hello! The answer "6" is wrong. This could be 6 if it was e^((ln(3)+ln(2)).
You can write e^((ln(3)*ln(2)) as:

3^(ln(2))
or

2^(ln(3))

So, the final result is: 2.1414860639032773

Dms

yes if u plug it into a calculator thats what you will obtain however math_lover101 is asking for the exact exponent as previously stated

juicybebe101:

yes if u plug it into a calculator thats what you will obtain however math_lover101 is asking for the exact exponent as previously stated

Yes, but its:

How do you find the exponent for this problem "e^((ln(3)*ln(2))"?

I think that there might be some confusion over this problem.

Let's look at a "log" rule:

We have
log(a*b) = log a + log b

But we DON'T have any specific "rule" for log(a) * log (b). In essence, we just "plug" ln(2)*ln(3) straight into a calculator to find the exponent!!

Note...if we had e^(lh(2) + ln(3)), this would mean the same thing as e^(ln(2))*e^(ln(3)) and THAT answer Is = "6"......This is based on the rule that e^(ln(a)) = a

I hope this helps.

math_lover101:

How do you find the exponent for this problem "e^((ln(3)*ln(2))"?

I will say the same as the CPhill and Dms, (oh, juicybebe, do you agree now?)