Chapter 6.6, Problem 72E

To determine

To calculate: .The equation for the value of $x$ .

Answer to Problem 72E

The required value of $x$ is $1$ .

Explanation of Solution

Given Information:

The given equation is $5^{2x}+20\cdot 5^x-125=0$ .

Formula Used:

Power rule of logarithmic functions.

  ${\log }_a{x}^p=p{\log }_{a}x$

Where $a$ is the base, $x$ is the variable and $p$ is the power of variable.

Property of logarithm.

  ${\log }_{a}a=1$

Quadratic formula.

If $a{x}^2+bx+c=0$ then $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

Where $a,b,c$ are constant and $a\ne 0$

Calculation:

Consider the equation is $5^{2x}+20\cdot 5^x-125=0$ .

Substitute $y$ for $5^x$ in the above equation.

  $y^2+20y-125=0$

Use quadratic formula of the above equation.

  $y=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

Substitute $1$ for $a$ , $20$ for $b$ and $-125$ for $c$ in the above formula.

  $y=\frac{-20\pm \sqrt{{\left(20\right)}^2-4\left(1\right)\left(-125\right)}}{2\left(1\right)}$

Simplify the above equation.

  $\begin{array}{c}y=\frac{-20\pm \sqrt{{\left(20\right)}^2-4\left(1\right)\left(-125\right)}}{2\left(1\right)}\\ =\frac{-20\pm \sqrt{400+500}}{2}\\ =\frac{-20\pm \sqrt{900}}{2}\\ =\frac{-20\pm 30}{2}\end{array}$

Simplify the equation.

  $\begin{array}{c}y=\frac{-20\pm 30}{2}\\ =\frac{-20+30}{2},\frac{-20-30}{2}\\ =\frac{10}{2}.\frac{-50}{2}\\ =5,-25\end{array}$

Now, $y=5$

Substitute $5^x$ for $y$ in the above result.

  $5^x=5$

Take logarithm each side with base $5$ .

  ${\log }_5{5}^x={\log }_{5}5$

Use power rule of logarithm ${\log }_a{x}^p=p{\log }_{a}x$ in the above equation.

  $x{\log }_{5}5={\log }_{5}5$

Use property of logarithm ${\log }_{a}a=1$ in the above equation.

  $\begin{array}{c}x\left(1\right)=1\\ x=1\end{array}$

And

  $y=-25$

Substitute $5^x$ for $y$ in the above result.

  $5^x=-25$

Take logarithm each side with base $5$ .

  ${\log }_5{5}^x={\log }_5\left(-25\right)$

It is known that ${\log }_a\left(-x\right)$ is does not exist.

Hence, the required value of $x$ is $1$ .