# Question cab16

Dec 8, 2017

$x \approx 0.78$

#### Explanation:

Ok before I begin, I would like to point out that $\log 6$ ACTUALLY means ${\log}_{10} 6$. When the subscript is not stated, it is implied to be $10$.

Ok, so logarithms are simply the inverse to exponentials. An easy way to ALWAYS solve logarithms easily is just remember this simple rule.

${\log}_{a} b = c$ is the same as $b = {a}^{c}$

Heres an example:

${\log}_{2} 64 = 6$ OR $64 = {2}^{6}$

log_10 6 = ? OR 6=10^?#

Honestly this is a very difficult problem to solve without a calculator. My calculator says that $x \approx 0.78$

~Chandler Dowd

Dec 8, 2017

$\log \left(6\right) \cong 0.778$

#### Explanation:

Logarithms can be thought of as the reverse of exponents. The expression ${\log}_{b} \left(x\right) = y$ is the same as solving the following equation for $y$:

${b}^{y} = x$

So $\log \left(6\right)$ can be thought of "what number do I have to raise $10$ to to get $6$?" ($\log$ is usually shorthand for ${\log}_{10}$).

In this case, we only have the values of the logarithms provided, so we're going to take advantage of the fact that $6$ can be written as $3 \cdot 2$

$\log \left(6\right) = \log \left(3 \cdot 2\right)$

Then I'm going to take advantage of the following logarithm property:
${\log}_{x} \left(a \cdot b\right) = {\log}_{x} \left(a\right) + {\log}_{x} \left(b\right)$

So our logarithm becomes:
$\log \left(3 \cdot 2\right) = \log \left(3\right) + \log \left(2\right) \cong 0.301 + 0.477 = 0.778$