logarithms with exponents

Question

I am supposed to solve for $x$ when $\log_{10}\big((x+1)^2\big) = 2.$ I used the log power rule and then got $2\cdot \log_{10}(x+1) = 2.$ When I graph both these functions, I get two different graphs. Why does the power rule not work here? And how should I attack this problem?

Answer 1

If $a$ is a non-zero real number, then $\log_{10}(a^2)=2\log_{10}\left(\color{red}{|a|}\right)$. (The "rule" $\log_{10}(a^2)=2\log_{10}(a)$ is actually only valid when $a$ is positive.) Applying this to the problem at hand, we get $$ \log_{10}\left((x+1)^2\right)=2\log_{10}\left(|x+1|\right)=2\implies \log_{10}\left(|x+1|\right)=1 \, . $$ By the definition of logarithms, this means that $|x+1|=10^1$. If $x+1\ge0$, then this equation becomes $x+1=10$ and so $x=9$. If $x+1<0$, then the equation becomes $-(x+1)=10$ and so $x=-11$.

Answer 2

Note that$$\log_{10}\bigl((x+1)^2\bigr)=\log_{10}\bigl(|x+1|^2\bigr)=2\log_{10}|x+1|.$$And $\log_{10}|x+1|$ is defined for every number other than $-1$, just as $\log_{10}\bigl((x+1)^2\bigr)$, and unlike $\log_{10}(x+1)$.

Answer 3

You can bring to common base: $$\log_{10} \left((x+1)^2\right)=2 \iff \log_{10}\left((x+1)^2\right)=\log_{10}10^2 \iff \\ (x+1)^2=100 \iff x+1=\pm 10 \iff x_1=-11, x_2=9.$$