Simplify log ad graph exponential

**nerdzor** · May 16th 2009, 09:59 PM

Simplify $\log a^3 - \log a^2$

I'm not sure if this is right but is it either:

a) $= 3\log_{10} a - 2\log_{10} a$
$=\log_{10} a (3-2)$
$=\log_{10} a {1}$
$a=0$

OR

b) $\log a^3 - \log a^2$
$=\log_{10} \frac{a^3}{a^2}$
$=\log_{10} a$

???

To graph $y=e^x$
$-1 \leq x\leq 3$
Is it just the exact same as a normal $y=e^x$ but asymptotic at $x=-1$ and $x=3$ ?

How can I find the gradient of the tangent at (0,1)?

**mathaddict** · May 16th 2009, 11:27 PM

Originally Posted by nerdzor

Simplify $\log a^3 - \log a^2$

I'm not sure if this is right but is it either:

a) $= 3\log_{10} a - 2\log_{10} a$
$=\log_{10} a (3-2)$
$=\log_{10} a {1}$
$a=0$

OR

b) $\log a^3 - \log a^2$
$=\log_{10} \frac{a^3}{a^2}$
$=\log_{10} a$

b is right .

**Singaporean** · May 16th 2009, 11:55 PM

b is right
because from your method A

if you substitute log a = b
it would be 3b-2b = b = log a

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