After the exam, a group of students checked their solutions and Adam wrote his solution on the board. Their task was to solve a logarithmic inequality: $$ \log x-1 \geq 0 $$ Here is Adam’s solution:
(1) Adam realized that logarithms can only be taken for positive numbers. Therefore, he added a solvability condition to the given inequality: $$ \log x-1 \geq 0;x>0 $$
(2) He converted the number $0$ on the right side of the inequality to $\log 1$: $$ \log x-1 \geq \log 1 $$
(3) He removed logarithms from both sides of the inequality: $$ x-1 \geq 1 $$
(4) He solved the resulting inequality and wrote the result using an interval: $$ \begin{gather} x \geq 2 \cr x \in [ 2;\infty) \end{gather} $$
(5) Finally, Adam did a check by choosing an arbitrary number from the above interval. Check for $x=100$: $$ \begin{gather} L=\log 100-1=2-1=1 \cr R=0 \cr L>R \end{gather} $$ Is Adam's solution correct? Explain.
No. The mistake is in step (3). In this form, it is not possible to remove logarithms from our inequality because the number $-1$ is not part of the logarithm on the left side.
No. The mistake is in step (2). We cannot write $0=\log 1$. It is a logarithm with a base of $10$ and thus it should be $0=\log 10$.
No. The mistake is in step (3). When removing logarithms, the sign of the inequality should have changed.
Yes. Adam's solution is correct.
Adam correctly determined the condition for solving but made a mistake when removing the logarithms from the inequality. The correct procedure of solving the inequality should have been:
Setting solvability conditions (logarithms can only be taken for positive numbers): $$ \log {x}-1 \geq 0;~x>0 $$ Adding one to both sides of the inequality: $$ \log x≥1 $$ Rewriting the right side using the logarithm: $$ \log {x} \geq \log 10 $$ Removing logarithms from the inequality: $$ x \geq 10 $$ The solution set is the interval $[ 10; \infty )$.
Note: The solution set of our inequality is an interval and the check (step (5)) cannot be performed by substituting only one selected number from this interval. In this case, the check is not necessary, and we do not have to do it at all.