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How do you find the exponent for this problem "e^((ln(3)*ln(2))"?
 Mar 14, 2014
 #1
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its 6 you multiply the two powers
 Mar 14, 2014
 #2
avatar+64 
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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
 Mar 14, 2014
 #3
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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
 Mar 14, 2014
 #4
avatar+64 
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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:
0
e^(ln(6))
So, the only solutions are: 3^(ln(2)) AND 2^(ln(3))

Dms
 Mar 14, 2014
 #5
avatar+126661 
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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.
 Mar 14, 2014
 #6
avatar+118579 
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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?)

e^((ln(3)*ln(2))

OR

e ln(3)*ln(2) ==> {e ln(3)}^[ln(2)] ==> 3 ln2 ==> 2.14148.....

OR

e ln(2)*ln(3) ==> {e ln(2)}^[ln(3)] ==> 2 ln3 ==> 2.14148.....

Now I think that this is an irrational number which means that no matter how many digits you express it to it will still be an approximation
 Mar 15, 2014

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