Chapter 3.5, Problem 25E

Solution

To solve: the given equation by two methods.

  $x=\pm 1000$

Given information:

The equation:

  $\log x^2=6$

Definition Used:

Common logarithm- Base 10:

The common logarithmic function ${\log }_{10}x=\log x$ is the inverse of the exponential function $f\left(x\right)={10}^x$ . Therefore,

  $y=\log x$ if and only if ${10}^y=x$

Property used:

One-to-One Properties

For any exponential function $f\left(x\right)=b^x$ ,

If $b^u=b^v$, then $u=v$ .

For any logarithmic function $f\left(x\right)={\log }_{b}x$ ,

If ${\log }_{b}u={\log }_{b}v$, then $u=v$ .

Explanation:

First Method: Changing logarithm to exponential form:

By the definition of common logarithm:

  $y=\log x$ if and only if ${10}^y=x$

Applying this to the given equation:

  $\log x^2=6$ if and only if ${10}^6=x^2$

That is,

  $\begin{array}{c}{x}^2={10}^6\\ x=\pm \sqrt{{10}^6}\\ x=\pm {10}^3\\ x=\pm 1000\end{array}$

Second method: Use the one-to-one property of logarithms.

  $\begin{array}{c}\log x^2=6\\ \log x^2=\log {10}^6\\ x^2={10}^6\\ x=\pm 1000\end{array}$