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Chapter 3: Problem 5

Determine whether each \(x\) -value is a solution (or an approximate solution) of the equation. \(\log _{2}(x+3)=10\) (a) \(x=1021\) (b) \(x=17\) (c) \(x=10^{2}-3\)

Short Answer

Expert verified
Only \(x=1021\) is a solution to the equation \(\log _{2}(x+3)=10\).

Step by step solution

01

Understand the equation

The equation provided: \(\log _{2}(x+3)=10\) needs to be solved using the given values of \(x\) which are 1021, 17 and \(10^{2}-3\).
02

Substitution for \(x=1021\)

Substitute \(x=1021\) into the equation:\[\log_{2}(1021+3)=10\]Simplifying it:\[\log_{2}(1024)=10\]Since \(2^{10}\) is 1024, the equality is true. Thus \(x=1021\) is a solution of the equation.
03

Substitution for \(x=17\)

Substitute \(x=17\) into the equation:\[\log_{2}(17+3)=10\]Simplify it:\[\log_{2}(20)=10\]However, \(2^{10}\) is not equal to 20, so \(x=17\) is not a solution.
04

Substitution for \(x=10^{2}-3\)

For the last case, substitute \(x=10^{2}-3\) into the equation:\[\log_{2}({10^{2}-3+3})=10\]Simplifying, we get \[\log_{2}({10^{2}}) =10 \]For \(2^{10} = 1024\), the equality is false because it is not equal to \(10^{2}\), hence \(x=10^{2}-3\) is not a solution.

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