Question

How do you solve ${{\log }_{2}}\left( 3x+1 \right)+{{\log }_{2}}\left( x+7 \right)=5$?

Answer 1

Hint: Now first we will simplify the equation using the property ${{\log }_{a}}\left( xy \right)={{\log }_{a}}x+{{\log }_{a}}y$ then we will convert the equation in exponent form and hence solve the obtained quadratic using the formula for quadratic equation $\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ .

Complete step-by-step answer:
Now consider ${{\log }_{2}}\left( 3x+1 \right)+{{\log }_{2}}\left( x+7 \right)=5$
The given equation is an equation in logarithms.
Now first we will simplify the given equation using the property of logarithm which states that ${{\log }_{a}}\left( mn \right)={{\log }_{a}}m+{{\log }_{a}}n$ .
Hence we can write the given equation as ${{\log }_{2}}\left[ \left( 3x+1 \right)\left( x+7 \right) \right]=5$ .
Now let us convert the equation in exponential form. We know that ${{\log }_{a}}x=n$ then we can write it as ${{a}^{n}}=x$ . Hence using this we get the given equation as,
$\begin{align}
  & \Rightarrow \left( 3x+1 \right)\left( x+7 \right)={{2}^{5}} \\
 & \Rightarrow \left( 3x+1 \right)\left( x+7 \right)=32 \\
\end{align}$
Now we know that according distributive property we have $\left( a \right)\left( b+c \right)=ab+ac$
Hence using this we get,
$\Rightarrow \left( 3x+1 \right)\left( x \right)+\left( 3x+1 \right)\left( 7 \right)=32$
Now again using distributive property we get,
$\Rightarrow 3{{x}^{2}}+x+21x+7=32$
Rearranging the terms in the equation we get,
$\Rightarrow 3{{x}^{2}}+22x+7-32=0$
$\Rightarrow 3{{x}^{2}}+22x-25=0$
Now the given equation is a quadratic equation in x.
Now let us use the formula $\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ to find the roots of the equation.
Now comparing the equation with the general form we get, a = 3, b = 22 and c = -25.
Now substituting the values in the formula we get,
$\begin{align}
  & \Rightarrow x=\dfrac{-22\pm \sqrt{{{22}^{2}}-4\left( 3 \right)\left( -25 \right)}}{2\left( 3 \right)} \\
 & \Rightarrow x=\dfrac{-22\pm \sqrt{484+300}}{6} \\
 & \Rightarrow x=\dfrac{-22\pm 28}{6} \\
\end{align}$
Hence the roots of the quadratic are $x=\dfrac{-22+28}{6}=1$ and $x=\dfrac{-22-28}{6}=\dfrac{-50}{6}$
Hence the solution of the given equation are x = 1 and $x=\dfrac{-25}{3}$

Note: Now note that we have ${{\log }_{a}}\left( mn \right)={{\log }_{a}}m+{{\log }_{a}}n$ and not ${{\log }_{a}}\left( m+n \right)={{\log }_{a}}m\times {{\log }_{a}}n$ note not to be confused in the formula for logarithm. Similarly we have ${{\log }_{a}}\left( \dfrac{m}{n} \right)={{\log }_{a}}m-{{\log }_{a}}n$ .