Question

Solve for the value (s) of x: - ${{\log }_{10}}5+{{\log }_{10}}\left( x+10 \right)-1={{\log }_{10}}\left( 21x-20 \right)-{{\log }_{10}}\left( 2x-1 \right)$?

Answer 1

Hint: First of all convert the constant 1 in the L.H.S by changing it into the logarithmic term with base 10 given as $1={{\log }_{10}}10$. Now, use the properties of log given as $\log m+\log n=\log \left( m\times n \right)$ and $\log m-\log n=\log \left( \dfrac{m}{n} \right)$ to simplify the L.H.S and R.H.S. Remove the log function from both the sides and equate their arguments and use the middle term split method to solve the obtained quadratic equation. Reject the values of x which make the argument of any logarithmic term equal to or less than 0.

Complete step-by-step answer:
Here we have been provided with the logarithmic expression ${{\log }_{10}}5+{{\log }_{10}}\left( x+10 \right)-1={{\log }_{10}}\left( 21x-20 \right)-{{\log }_{10}}\left( 2x-1 \right)$ and we are asked to solve for the values of x. Let us use some of the basic properties of log to get the answer.
Now, we know that if the base and the argument of the log are equal then its value is 1, so we can write 1 as ${{\log }_{10}}10$. Therefore the expression becomes: -
$\Rightarrow {{\log }_{10}}5+{{\log }_{10}}\left( x+10 \right)-{{\log }_{10}}10={{\log }_{10}}\left( 21x-20 \right)-{{\log }_{10}}\left( 2x-1 \right)$
Using the formulas $\log m+\log n=\log \left( m\times n \right)$ and $\log m-\log n=\log \left( \dfrac{m}{n} \right)$ to simplify the L.H.S and R.H.S we get,
$\begin{align}
  & \Rightarrow {{\log }_{10}}\left( \dfrac{5\times \left( x+10 \right)}{10} \right)={{\log }_{10}}\left( \dfrac{21x-20}{2x-1} \right) \\
 & \Rightarrow {{\log }_{10}}\left( \dfrac{x+10}{2} \right)={{\log }_{10}}\left( \dfrac{21x-20}{2x-1} \right) \\
\end{align}$
We can see that the base of the log function on both the sides is same so we can remove the log function and equate their arguments, so we get,
$\Rightarrow \left( \dfrac{x+10}{2} \right)=\left( \dfrac{21x-20}{2x-1} \right)$
By cross multiplication we get,
$\begin{align}
  & \Rightarrow \left( x+10 \right)\left( 2x-1 \right)=\left( 2 \right)\left( 21x-20 \right) \\
 & \Rightarrow 2{{x}^{2}}+19x-10=42x-40 \\
 & \Rightarrow 2{{x}^{2}}-23x+30=0 \\
\end{align}$
Using the middle term split method to factorize the above quadratic equation we get,
$\begin{align}
  & \Rightarrow 2{{x}^{2}}-20x-3x+30=0 \\
 & \Rightarrow 2x\left( x-10 \right)-3\left( x-10 \right)=0 \\
 & \Rightarrow \left( 2x-3 \right)\left( x-10 \right)=0 \\
\end{align}$
Substituting each term equal to 0 we get,
$\Rightarrow 2x-3=0$ or $x-10=0$
$\Rightarrow x=\dfrac{3}{2}$ or $x=10$
Clearly we can see that none of the above two values of x will make the argument of any logarithmic term equal to or less than 0, therefore both the values of x are valid.
Hence, the solutions of the given logarithmic equation are $\dfrac{3}{2}$ and 10.

Note: Note that you must not remove the log function from all the terms in the initial step of the solution. First we have to simplify both the sides so that they contain a single log function on both the sides and then only we can compare the argument and the condition that must be full filled is that they must have the same base. Do not forget to check the values of x obtained because log is undefined for non – positive arguments.