Question

Evaluate the following equation
$$\log \left(x+3\right)+\log \left(x-3\right)=\log 16$$

Answer 1

Hint: In order to solve the problem first simplify the terms by the use of logarithmic identities. Try to separate the unknown variable term and proceed to find the answer.

Complete step-by-step answer:
Given equation is $$\log \left(x+3\right)+\log \left(x-3\right)=\log 16$$
We know the identity for sum of the logarithmic terms
$\log a+\log b=\log \left(ab\right)$
Using the above formula let us simplify the LHS
$$\begin{gathered} \Rightarrow \log \left(x+3\right)+\log \left(x-3\right)=\log 16 \\ \Rightarrow \log \left[\left(x+3\right)\left(x-3\right)\right]=\log 16 \end{gathered}$$
Now we have logarithmic terms on both sides.
As we know the general rule of logarithm which is
$$\begin{gathered} \text{if }\log c=\log d \\ \Rightarrow c=d \end{gathered}$$
Using the same rule in above equation we get
$$\begin{gathered} \because \log \left[\left(x+3\right)\left(x-3\right)\right]=\log 16 \\ \Rightarrow \left(x+3\right)\left(x-3\right)=16 \end{gathered}$$
Now in order to find the value of x we need to solve the above algebraic equation
$$\begin{gathered} \Rightarrow \left(x+3\right)\left(x-3\right)=16 \\ \Rightarrow x^2-3^2=16\left[\because \left(a+b\right)\left(a-b\right)=a^2-b^2\right] \\ \Rightarrow x^2-9=16 \\ \Rightarrow x^2=16+9=25 \\ \Rightarrow x^2-25=0 \\ \Rightarrow x^2-5^2=0 \\ \Rightarrow \left(x+5\right)\left(x-5\right)=0 \\ \Rightarrow x=5\mathrm{\&}x=-5 \end{gathered}$$
So after solving the algebraic equation we have 2 values of x which are 5 and -5 but when we substitute the value -5 in the given term the value of term becomes $\log \left(-5+3\right)=\log \left(-2\right)$ , but the log of negative number does not exist so +5 is the only solution.
Hence, the value of x is 5.

Note: The logarithm is the inverse function to exponentiation. That means the logarithm of a given number x is the exponent to which another fixed number, the base b, must be raised, to produce that number x. Students must remember that logarithm of negative numbers does not exist but logarithm of some positive numbers can be negative.