Question

Solve the following equation:
$\log \left( {\log x} \right) + \log \left( {\log {x^3} - 2} \right) = 0$

Answer 1

Hint: Use the properties of log to simplify the formula. You can see that in both subparts, log is inside the log so we need to make them one so that at least we have some homogeneous equation to work with. After doing that, we can remove the first log, one which is outside the box to further simplify and solve the equation.

Formulas used:
Multiplication of additives property
$\log a + \log b = \log ab$
Structure representation of log
$\log a = b \Rightarrow a = {10^b}$
Power property
$\log {a^b} = b \cdot \log a$

Complete step by step solution:
In this question we have to work with the equation $\log \left( {\log x} \right) + \log \left( {\log {x^3} - 2} \right) = 0$. We know that $\log a + \log b = \log ab$, so the equation will be
 $\
  \log \left( {\log x} \right) + \log \left( {\log {x^3} - 2} \right) = 0 \\
  \log \left( {\log x.\left( {\log {x^3} - 2} \right)} \right) = 0 \\
\ $
We also know that $\log a = b \Rightarrow a = {10^b}$, so the equation can further be simplified as
$\
  \log \left( {\log x.\left( {\log {x^3} - 2} \right)} \right) = 0 \\
  \log x.\left( {\log {x^3} - 2} \right) = {10^0} \\
  \log x.\left( {\log {x^3} - 2} \right) = 1 \\
  \log x.\log {x^3} - 2\log x = 1 \\
  3\log x \cdot \log x - 2\log x = 1 \\
\ $
In the last step power property is used. Now, let $\log x = k$. We can further simplify this equation as
$\
  3\log x \cdot \log x - 2\log x = 1 \\
  3k \cdot k - 2k = 1 \\
  3{k^2} - 2k = 1 \\
  3{k^2} - 2k - 1 = 0 \\
\ $
Now we simply need to solve this quadratic equation to find the value of k. for this we add and subtract k in the left-hand side. Thereafter, we will take commons.
$\
  3{k^2} - 2k - 1 = 0 \\
  3{k^2} - 2k - 1 + k - k = 0 \\
  3{k^2} - 2k - k + k - 1 = 0 \\
  3{k^2} - 3k + k - 1 = 0 \\
  3k\left( {k - 1} \right) + 1\left( {k - 1} \right) = 0 \\
  \left( {3k + 1} \right)\left( {k - 1} \right) = 0 \\
\ $
Thus either $k - 1 = 0$ or $3k + 1 = 0$ which means that k is either 1 or $\dfrac{{ - 1}}{3}$
But we assumed $\log x = k$, so either
$\
  \log x = 1 \\
  x = {10^1} = 10 \\
\ $
Or,
$\
  \log x = \dfrac{{ - 1}}{3} \\
  x = {10^{\dfrac{{ - 1}}{3}}} \\
\ $

Note:
Remember, this addition and subtraction will not affect equality because we are not increasing anything on any side, rather just adding zero to one side which will not change it.
Also, when working with logarithmic functions, remember to keep an eye out for ln because log has base 10 and ln has base ‘e’ which is about 2.7183 and if you confuse them, you will end up making blunders.