Question

How does one solve $\log {x^3} + \log 8 = 3$?

Answer 1

Hint:For simplifying the original equation , firstly used logarithm property $\log a + \log b = \log (ab)$ then take base ten exponential of both sides of the equation, then apply the logarithm formula ${b^{{{\log }_b}a}} = a$ to simplify the equation and lastly take cube root both sides.

Formula used:
We used logarithm properties i.e.
$\log a + \log b = \log (ab)$ ,
And
${b^{{{\log }_b}a}} = a$ , the logarithm function says $\log x$ is only defined when $x$ is greater than zero.


Complete solution step by step:
It is given that ,
$
\log {x^3} + \log 8 = 3 \\
or \\
\log (8{x^3}) = 3 \\
$
Since we know logarithm properties i.e.
$\log a + \log b = \log (ab)$ ,
Now, by assuming the base of the logarithm to be ten ,then take the base ten exponential of both sides of the equation, we will get the following result ,
${10^{{{\log }_{10}}(8{x^3})}} = {10^3}$
By applying the logarithm formula ${b^{{{\log }_b}a}} = a$ . we will get ,
$(8{x^3}) = 1000$
 Simplify the equation, we will the following result ,
$ \Rightarrow {(2x)^3} = {10^3}$
Taking cube root both the side , we will get ,
$
\Rightarrow 2x = 10 \\
\Rightarrow x = 5 \\
$
Now recall that the logarithm function says $\log x$ is only defined when $x$is greater than zero.
Therefore, in our original equation $\log {x^3} + \log 8 = 3$ ,
Here,
$({x^3}) > 0$ ,
For $x = 5$ ,
${5^3} > 0$
Therefore, we have our solution i.e., $5$ .

Note: The logarithm function says $\log x$ is only defined when $x$ is greater than zero. While defining logarithm function one should remember that the base of the log must be a positive real number and not equals to one . At the end we must recall that the logarithm function says $\log x$ is only defined when $x$is greater than zero. While performing logarithm properties we have
remember certain conditions , our end result must satisfy domain of that logarithm