# How do you combine \log _ { 9} 36- \log _ { 3} 2+ \log _ { 0.5} 4?

Aug 28, 2017

$- 1$

#### Explanation:

So what the logarithm function is asking is, what is the exponent that the base needs to be raised to to obtain the value, ie

${\log}_{w} \left(z\right) = v$

Means that ${w}^{v} = z$

So what we have is that

${9}^{a} = 36$, ${3}^{b} = 2$ and ${0.5}^{c} = 4$

Take logs of both sides:

$\log \left({9}^{a}\right) = \log \left(36\right)$, $\log \left({3}^{b}\right) = \log \left(2\right)$ and $\log \left({0.5}^{c}\right) = \log \left(4\right)$

Now use the exponent rule of logs, ie

$\log \left({w}^{z}\right) = z \log \left(w\right)$

$\therefore a = \log \frac{36}{\log} \left(9\right) , b = \log \frac{2}{\log} \left(3\right) \mathmr{and} c = \log \frac{4}{\log} \left(0.5\right)$

$a = \log \frac{36}{\log} \left(9\right) = \log \frac{{6}^{2}}{\log} \left({3}^{2}\right) = \frac{2 \cdot \log \left(6\right)}{2 \cdot \log \left(3\right)} = \log \frac{6}{\log} \left(3\right)$

Since we're going to be using b as well it would be helpful if we could get a into a similar form...luckily we can! Using the addition rule of logs:

$\log \left(w\right) + \log \left(z\right) = \log \left(w z\right)$

So:

$a = \log \frac{6}{\log} \left(3\right) = \log \frac{2 \cdot 3}{\log} \left(3\right) = \frac{\log \left(2\right) + \log \left(3\right)}{\log} \left(3\right) = 1 + \log \frac{2}{\log} \left(3\right)$

That's us got a and b looking nice so now we turn our attention to c.

$c = \log \frac{4}{\log} \left(0.5\right) = \log \frac{{2}^{2}}{\log} \left({2}^{-} 1\right) = - 2$

Therefore

$a - b + c = 1 + \log \frac{2}{\log} \left(3\right) - \log \frac{2}{\log} \left(3\right) - 2 = - 1$