value of log

**dm10** · September 9th 2008, 04:29 PM

(a) e ^ -2ln5

(b)ln(ln e^e^10)

The second one is e to the e to the tenth power.

**Prove It** · September 9th 2008, 04:37 PM

Originally Posted by dm10

(a) e ^ -2ln5

(b)ln(ln e^e^10)

The second one is e to the e to the tenth power.

For a) Do you remember the rule $\log_a(m^p)=p\log_a(m)$ ?

Using this rule, we get

$e^{-2\ln(5)}=e^{\ln(5^{-2})}=5^{-2}=\frac{1}{5^2}=\frac{1}{25}$

The second actually just reduces to 10, because $\ln (e^{e^{10}})=e^{10}$ and so
$\ln[\ln(e^{e^{10}})]=\ln[e^{10}]=10$ .

**dm10** · September 9th 2008, 04:40 PM

Originally Posted by Prove It

For a) Do you remember the rule $\log_a(m^p)=p\log_a(m)$ ?

Using this rule, we get

$e^{-2\ln(5)}=e^{\ln(5^{-2})}=5^{-2}=\frac{1}{5^2}=\frac{1}{25}$

The second actually just reduces to 10, because $\ln(e^e^10)=e^10$ and so
$\ln[\ln(e^e^10)]=\ln[e^10]=10$ .

Thank you, you're answers really help me understand the material.

**Prove It** · September 9th 2008, 04:42 PM

Originally Posted by dm10

Thank you, you're answers really help me understand the material.

No problem, and I just fixed the LaTeX errors in my post.

**dm10** · September 9th 2008, 04:56 PM

Originally Posted by Prove It

No problem, and I just fixed the LaTeX errors in my post.

I've got one more left if you don't mind helping:

arcsin 2

Find the exact value.

**mr fantastic** · September 9th 2008, 04:58 PM

Originally Posted by dm10

I've got one more left if you don't mind helping:

arcsin 2

Find the exact value.

Do you realise that arcsin 2 has no real value?

**dm10** · September 9th 2008, 05:00 PM

Originally Posted by mr fantastic

Do you realise that arcsin 2 has no real value?

So the answer would be no real value?

**mr fantastic** · September 9th 2008, 05:04 PM

Originally Posted by dm10

So the answer would be no real value?

If you're working exclusively with real numbers it would be reasonable to say that arcsin(2) is not defined.

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