Peter had to solve a logarithmic equation: $$ (\log{x} )^2+9 \log{x}=0 $$ After handing it in, the teacher told him that he had made a serious mistake in his solution and asked him to find it. Let's look at his solution and try to identify the error:
(1) At first, Peter set the condition for $x$: $$ x>0 $$
(2) By introducing the substitution $\log{x}=t$, he obtained a quadratic equation: $$ t^2+9t=0 $$
(3) He divided the whole equation by $t$ and rearranged it to the form: $$ t+9=0 $$
(4) The solution is then: $$ t=-9 $$
(5) By substituting back to the substitution, he obtained the equation: $$ \log{x}=-9 $$ and using the definition of the logarithm, he obtained the solution: $$ x=10^{-9} $$
(6) Finally, Peter verified his solution by performing a check: $$ L=(\log{10^{-9} })^2+9 \log{10^{-9}}=(-9 \log{10})^2+9 \cdot (-9) \log{10}=81-81=0,~ $$ $$ R=0, ~ R=L $$ In which step did the error occur?
The error is in step (1). The correct solvability condition should be $x\geq0$.
The error is in step (3). By dividing by $t$, Peter lost one solution.
The error is in step (4). The root $–9$ is negative and therefore does not satisfy the solvability condition. This implies that the equation has no solution.
The error is in step (5). Equation $\log{x}=-9$ cannot be converted to $x=10^{-9}$.
The error is in step (6). The left side of the equation is incorrectly modified. It should be $L=81+81=162$.
The correct solution to the equation $$ (\log {x} )^2+9 \log{x}=0 $$ starts with determining the domain of the logarithm ($x>0$) and introducing the substitution $\log{x}=t$. After using the substitution, we obtain the (incomplete) quadratic equation $t^2+9t=0$, which can be solved by factoring $t$ out in the left side of the equation: $$ t(t+9)=0 $$ Now, let us realize that the product is equal to zero if and only if at least one of the factors is zero. That is, $t=0$ or $t+9=0$. So, the substitution roots are $t=0$ and $t=-9$.
Now, we return to the substitution $\log{x} =t$:
a) $t=0 \Rightarrow \log{x} =0 \Rightarrow x=1$
b) $t=-9 \Rightarrow \log{x} =-9 \Rightarrow x=10^{-9}$
The equation has two solutions $x=1$ and $x=10^{-9}$.