Answer
See explanation
Work Step by Step
**Answer Explanation**
Recall that by definition of the logarithm, \(\log_b(a)\) is “the exponent you put on \(b\) to get \(a\).” Formally, \(\log_b(a) = y\) means exactly that \(b^y = a\), assuming \(b>0\) and \(b \neq 1\).
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**(a) Why \(\log_b\bigl(b^x\bigr) = x\)**
1. Look at the expression \(\log_b\bigl(b^x\bigr)\). By definition, \(\log_b\bigl(b^x\bigr)\) is the exponent to which you raise \(b\) to obtain \(b^x\).
2. But we already know that raising \(b\) to the power \(x\) gives \(b^x\).
3. Therefore, the exponent in question must be \(x\). Hence \(\log_b\bigl(b^x\bigr) = x\).
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**(b) Why \(b^{\log_b(x)} = x\)**
1. By definition, \(\log_b(x)\) is the exponent that makes \(b\) become \(x\). In other words, \(\log_b(x) = y\) means \(b^y = x\).
2. If you take \(b\) raised to \(\log_b(x)\), you are literally “raising \(b\) to the exponent that gives \(x\).”
3. Hence \(b^{\log_b(x)} = x\).
Both parts are simply restatements of the fact that “logarithm and exponentiation (with the same base) are inverse operations.”
