Question

How do you solve the equation $\log \left( x+6 \right)=1-\log \left( x-5 \right)$

Answer 1

Hint: Now to simplify the equation we will first rearrange the terms of the equation by taking all log terms together. Then we will use the property $\log \left( ab \right)=\log a+\log b$ and simplify the equation. Now with the help of definition of log we will eliminate the log from the equation hence find the solution to the equation.

Complete step by step solution:
Now let us first understand the concept of logarithm.
Now consider a number ${{\log }_{a}}b$ where a is called the base of logarithm.
Now log gives us the exponent to which the base must be raised to get the required number.
Now let us say we have an equation ${{\log }_{a}}x=n$ then we have ${{a}^{n}}=x$
Hence we can convert the logarithm to exponent anytime by this method
Now note that if nothing is written in the base then the base is taken as 10. Hence we have log10 = 1.
Now let us understand the basic rules of logarithm.
Rule of multiplication: According to this rule we have $\log \left( ab \right)=\log a+\log b$
Rule of division: According to this rule we have $\log \left( \dfrac{a}{b} \right)=\log a-\log b$
Rule of exponent: According to this rule we have $\log {{\left( a \right)}^{n}}=n\log a$
Now consider the given expression $\log \left( x+6 \right)=1-\log \left( x-5 \right)$.
Rearranging the terms in the equation we get,
$\Rightarrow \log \left( x+6 \right)+\log \left( x-5 \right)=1$
Now using the multiplication rule of log we get,
$\Rightarrow \log \left[ \left( x+6 \right)\left( x-5 \right) \right]=1$
Now by using the definition of log we have,
$\Rightarrow \left( x+6 \right)\left( x-5 \right)={{10}^{1}}$
Now simplifying the equation we get
$\begin{align}
  & \Rightarrow {{x}^{2}}-5x+6x-30=10 \\
 & \Rightarrow {{x}^{2}}+x-40=0 \\
\end{align}$
Now we have a quadratic in x of the form $a{{x}^{2}}+bx+c$ and we know the solution of the equation is given by the formula $\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ .
$\begin{align}
  & \Rightarrow \dfrac{-1\pm \sqrt{{{1}^{2}}-4\left( -40 \right)}}{2\left( 1 \right)} \\
 & \Rightarrow \dfrac{-1\pm \sqrt{1+160}}{2} \\
 & \Rightarrow \dfrac{-1\pm \sqrt{161}}{2} \\
\end{align}$

Hence the solution of the given equation is $\dfrac{1-\sqrt{161}}{2}$ and $\dfrac{1+\sqrt{161}}{2}$

Note: Now note that when nothing is written in the base of log we take it as 10. Also if the base of log is e then the log is called natural logarithm and is denoted by ln. Now also note that the base of log is always positive integer not equal to 1 and log1 = 0.