Express ln (3squareroot(32)) in terms of ln 2 and ln 3

[abel muroi]

I have been scratching my head for a couple of minutes now because of this problem.



Express ln (³squareroot(32)) in terms of ln 2 and ln 3




I'm not sure how to start. I know the rules for logarithms and i've done many problems involving logs, but this one in particular is difficult.

Can anyone show me the STEPS that it takes to solve this problem? (I usually learn best when someone tells me the steps i need to take to solve this)

 

[ksdhart]

My very first step would be to note that what you're given is actually the following:

\(\displaystyle \ln ^3\left(\sqrt{32}\right)=\left[\ln \left(\sqrt{32}\right)\right]^3=\ln \left(\sqrt{32}\right)\cdot \ln \left(\sqrt{32}\right)\cdot \ln \left(\sqrt{32}\right)\)

As you proceed forward from here, I'd remember the exponent and logarithm rules that you've learned. Specifically, recall that \(\displaystyle \sqrt{x}=x^{0.5}\) and \(\displaystyle \ln \left(a^b\right)=b\cdot \ln \left(a\right)\). Plus, think about how you might factor 32. Can you express 32 solely in terms of the numbers 2 and 3? What does that mean for your logarithms?

 

[abel muroi]

My very first step would be to note that what you're given is actually the following:

\(\displaystyle \ln ^3\left(\sqrt{32}\right)=\left[\ln \left(\sqrt{32}\right)\right]^3=\ln \left(\sqrt{32}\right)\cdot \ln \left(\sqrt{32}\right)\cdot \ln \left(\sqrt{32}\right)\)

As you proceed forward from here, I'd remember the exponent and logarithm rules that you've learned. Specifically, recall that \(\displaystyle \sqrt{x}=x^{0.5}\) and \(\displaystyle \ln \left(a^b\right)=b\cdot \ln \left(a\right)\). Plus, think about how you might factor 32. Can you express 32 solely in terms of the numbers 2 and 3? What does that mean for your logarithms?

actually...

The exponent is part of the square root, it is not part of the ln.

Im not sure how to type that...

³squareroot(32)

 

[Ishuda]

I have been scratching my head for a couple of minutes now because of this problem.



Express ln ³squareroot(32) in terms of ln 2 and ln 3




I'm not sure how to start. I know the rules for logarithms and i've done many problems involving logs, but this one in particular is difficult.

Can anyone show me the STEPS that it takes to solve this problem? (I usually learn best when someone tells me the steps i need to take to solve this)

Taking a different tack [because of the ln(3) part], I'm going to guess what you mean is
log₃[(32)^1/2]
That is the log base 3 of (32)^1/2. Noting that 32=2ⁿ for some n, you can write that as
A log₃(2)
for some A. What are the rules for changing bases [in this case from base 3 to base e (the natural log)]?

 

[abel muroi]

Taking a different tack [because of the ln(3) part], I'm going to guess what you mean is
log₃[(32)^1/2]
That is the log base 3 of (32)^1/2. Noting that 32=2ⁿ for some n, you can write that as
A log₃(2)
for some A. What are the rules for changing bases [in this case from base 3 to base e (the natural log)]?

Sorry, I meant to type ln (³squareroot(32)), not ln ³Squareroot(32)

 

[Ishuda]

Sorry, I meant to type ln (³squareroot(32)), not ln ³Squareroot(32)

Do you mean the cube root of 32, i.e. \(\displaystyle \sqrt[3]{32}\) or \(\displaystyle 32^{\frac{1}{3}}\)

 

[pka]

Sorry, I meant to type ln (³squareroot(32)), not ln ³Squareroot(32)

\(\displaystyle \Large\log(\sqrt[3]{32})=\frac{5}{3}\log(2)\)

 

[abel muroi]

\(\displaystyle \Large\log(\sqrt[3]{32})=\frac{5}{3}\log(2)\)

But how do i express ln (³squareroot(32)) in terms of ln 2 and ln 3?

 

[pka]

The exponent is part of the square root, it is not part of the ln.
³squareroot(32)

But how do i express ln (³squareroot(32)) in terms of ln 2 and ln 3?

Why the hel* will you not tell us exactly what ³squareroot(32) means?
If it is \(\displaystyle \Large\sqrt[3]{32}\) you type that as "the cube root of 32"
If it is that there is no \(\displaystyle \bf{\log(3)}\)

Now \(\displaystyle \log_3(32)=\dfrac{5\log(2)}{\log(3)}\), in which case type "log base 3 of 32"

So you don't dare blame any of us because you do not seem to know what you are doing.