Question

How do you solve ${\log }_2(3x+1)+{\log }_2(x+7)=5$?

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 ${\displaystyle \frac{-b\pm \sqrt{b^2-4ac}}{2a}}$ .

Complete step-by-step answer:
Now consider ${\log }_2(3x+1)+{\log }_2(x+7)=5$
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 ${\log }_2\left[(3x+1)(x+7)\right]=5$ .
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,
$\begin{array}{rl}& \Rightarrow (3x+1)(x+7)=2^5\\ & \Rightarrow (3x+1)(x+7)=32\end{array}$
Now we know that according distributive property we have $\left(a\right)(b+c)=ab+ac$
Hence using this we get,
$\Rightarrow (3x+1)\left(x\right)+(3x+1)\left(7\right)=32$
Now again using distributive property we get,
$\Rightarrow 3x^2+x+21x+7=32$
Rearranging the terms in the equation we get,
$\Rightarrow 3x^2+22x+7-32=0$
$\Rightarrow 3x^2+22x-25=0$
Now the given equation is a quadratic equation in x.
Now let us use the formula ${\displaystyle \frac{-b\pm \sqrt{b^2-4ac}}{2a}}$ 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,
$\begin{array}{rl}& \Rightarrow x={\displaystyle \frac{-22\pm \sqrt{{22}^2-4\left(3\right)(-25)}}{2\left(3\right)}}\\ & \Rightarrow x={\displaystyle \frac{-22\pm \sqrt{484+300}}{6}}\\ & \Rightarrow x={\displaystyle \frac{-22\pm 28}{6}}\end{array}$
Hence the roots of the quadratic are $x={\displaystyle \frac{-22+28}{6}}=1$ and $x={\displaystyle \frac{-22-28}{6}}={\displaystyle \frac{-50}{6}}$
Hence the solution of the given equation are x = 1 and $x={\displaystyle \frac{-25}{3}}$

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