Math CentralQuandaries & Queries


Question from Émile:

If you had 2^a = 4 and 2^b = 4 you could assume that a = b right? Yet we can observe that 1^1 and 1^0 both equal 1, yet 1 = 0 isn't true. So if given a log base 1 number 1 the would the answer be 1 AND 0?

Hi Émile,

You say "If you had $2^a = 4$ and $2^b = 4$ you could assume that $a = b$ right?" Suppose you didn't want to assume that $a = b$ but rather to prove it. How would you do it?

I would start with $2^a = 2^b$ and take the logarithm of both sides to get

\[\log\left(2^a\right) = log\left(2^b\right).\]

Using a property of the logarithm function this can be rewritten as

\[a \log(2) = b \log(2).\]

Dividing both sides by $\log(2)$ yields $a = b.$

Now apply the same argument to the equation $1^1 = 1^{0}.$ Taking the logarithm of both sides yields

\[\log\left(1^1\right) = log\left(1^0\right)\]


\[1 \log(1) = 0 \log(1).\]

But now I am stuck since $\log(1) = 0$ and thus I can't divide both sides by $\log(1).$ All the equation

\[1 \log(1) = 0 \log(1)\]

says is $0 = 0$ and it doesn't allow me to conclude that $0 = 1.$

I hope this helps,

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