Question

How do you solve the equation $\dfrac{1}{3}\log x = \log 8?$

Answer 1

Hint: Here first of all we will Apply Power rule: ${\log _a}{x^n} = n{\log _a}x$ and then will simplify the equation and then will find out the resultant required value for “x” using cubes and cube-root concepts.

Complete step-by-step solution:
Take the given expression: $\dfrac{1}{3}\log x = \log 8$
Apply Power rule: ${\log _a}{x^n} = n{\log _a}x$ in the above given equation.
$\log {x^{\dfrac{1}{3}}} = \log 8$
Now, log from both the sides of the equation cancel each other. Therefore, more simplified equation will be –
$ \Rightarrow {x^{\dfrac{1}{3}}} = 8$
Take a cube on both the sides of the equation.
$ \Rightarrow {\left( {{x^{\dfrac{1}{3}}}} \right)^3} = {8^3}$
Applying the law of power and exponent which states that the power to the power is multiplied.
$ \Rightarrow \left( {{x^{\dfrac{1}{3} \times 3}}} \right) = {8^3}$
Common factors from the numerator and the denominator cancel each other on the left hand side of the equation.
$ \Rightarrow x = {8^3}$
Simplify the above expression on the right hand side. Cube means the number multiplied with itself thrice.
$ \Rightarrow x = 8 \times 8 \times 8$
Simplify the above expression finding the product on the right hand side of the equation.
$ \Rightarrow x = 512$
Therefore this is the required answer.

Additional Information: Also refer to the below properties and rules of the logarithm.
Product rule: ${\log _a}xy = {\log _a}x + {\log _a}y$
Quotient rule: ${\log _a}\dfrac{x}{y} = {\log _a}x - {\log _a}y$
Power rule: ${\log _a}{x^n} = n{\log _a}x$
 Base rule:${\log _a}a = 1$
Change of base rule: ${\log _a}M = \dfrac{{\log M}}{{\log N}}$

Note: In other words, the logarithm is the power to which the number must be raised in order to get some other. Always remember the standard properties of the logarithm.... Product rule, quotient rule and the power rule. The basic logarithm properties are most important and the solution solely depends on it, so remember and understand its application properly. Be good in multiples and know the concepts of square and square root and apply accordingly.