The teacher chose three students, Peter, George, and John, to solve the equation: $$ 2^{x-1}=2-\log_22 $$ First, all three students modified the equation to the form: $$ 2^{x-1}=1 $$ Then, they proceeded differently:
Petr: He claimed that the equation has no solution. He reasoned that the value of a power of $2$ can never be equal to $1$.
George: He took the logarithm of both sides of the equation: $$ \begin{aligned} 2^{x-1} & =1 \cr \log 2^{x-1} & =\log 1 \end{aligned} $$ Then, he applied the logarithm rule: $$ \log_a x^n =n \cdot \log_a x $$ and obtained the equation: $$ (x-1) \log 2=\log 1 $$ He solved this equation as follows: $$ \begin{aligned} (x-1) \log2 & =\log 1 \cr x-1 & =\log \frac12 \cr x-1 & =\log 2^{-1} \cr x-1 & =-\log 2 \cr x & =1-\log 2 \end{aligned} $$
John: Realizing that the number $1$ can be expressed as $2^0$, he rewrote the equation as: $$ \begin{aligned} 2^{x-1} & =1 \cr 2^{x-1} & =2^0 \end{aligned} $$ Then, comparing exponents for the same base, he obtained the solution: $$ \begin{aligned} x-1 & =0 \cr x & =1 \end{aligned} $$ Whose procedure was correct?
John’s
Peter’s
George’s
Nobody’s. There is a mistake in each procedure.
John’s procedure is the correct one.
Peter made the mistake by claiming that the value of a power of $2$ can never be equal to $1$. He did not realize that $2^0=1$.
George made a mistake in the modification of the equation: $$ (x-1) \log 2=\log 1 $$ He specifically made the mistake when he tried to convert the expression $\log 2$ from the left side of the equation to the right side. It is not true that: $$\frac{\log 1 }{\log 2} =\log \frac12$$ If he had realized that $\log 1=0$, he would have gotten the equation: $$x-1=0$$ and thus the correct solution.