1. ## value of log

(a) e ^ -2ln5

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

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

2. 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
$\ln[\ln(e^{e^{10}})]=\ln[e^{10}]=10$.

3. 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.

4. 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.

5. 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.

6. 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?

7. Originally Posted by mr fantastic
Do you realise that arcsin 2 has no real value?
So the answer would be no real value?

8. 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.