Logarithms
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Solve 2^x = 32 for x.
Common mix-up
In this example, a student wrote x = 16.
The student divided 32 by 2 and answered 16, treating log2(32) as if a log asked for a quotient.
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See why this works
What power solves 2^x = 32?
A logarithm asks one question: what exponent do I need? log3(81) says: 3 to what power gives 81?
log base b of N is the exponent you raise b to in order to get N.
A logarithm is a question in disguise. log2(32) asks: 2 to what power equals 32? The answer is not a quotient. The log answers with an exponent, so dividing 32 by 2, as some do, finds a different number entirely and answers a question nobody asked.
To answer, rewrite the log as an exponential. log2(32) = x means exactly 2^x = 32. Then ask: how many 2's multiply together to make 32? Doubling along the way, 2, 4, 8, 16, 32, the count of multiplications is five, so 2^5 = 32 and x = 5.
The logarithm of 32 in base 2 is therefore 5: log2(32) = 5. Two expressions, one statement. When you see a log, always translate into exponential form first if the answer is not obvious. The exponential form shows where the unknown lives: it is an exponent, missing, and waiting to be counted.
This works cleanly whenever a number is a small power of the base. log3(81) asks: 3 to what power gives 81? Walking up: 3, 9, 27, 81, that is four multiplications, so the answer is 4. In general the log asks for the exponent, and the exponential form gives you a path to find it.
Keep this idea
A logarithm is an exponent. Ask which power, or rewrite the log in exponential form and count.
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