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I am doing a question on logarithms and am a bit confused regarding a solution I have found. As you can see below in the solution at one point the questions requires you to square (4ln(2))^2. When I read this this was my process:

  1. Square the 4 to 16 and take out of the bracket.
  2. I am left with ln(2)^2 so I move the power down using the log rule and do 16*2

See below:

enter image description here

But as you can see in the correct solution the person has left the ln(2)^2 and has not moved it down. Why is this? Am I not supposed to lower the power down using the log rule? I thought ln(2)^2 is the exact same as (ln(2))^2.

enter image description here

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2 Answers2

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Be careful about what carries the exponent. "$\ln(2^2)$" and "$(\ln 2)^2$" are not the same thing. In the first, you are squaring the $2$ then applying the logarithm, while in the second you are applying the logarithm and then squaring that.

In other words, the first is "$\ln(2\cdot 2)$", while the second is "$(\ln 2)\cdot(\ln 2)$".

The "log rule" applies to the first case, not the second.

Comment: Notice also how I have used parentheses at the start of my answer. It does not help to surround only the $2$ with parentheses as you have done, because it is still unclear what carries the exponent. It is usually understood, if there are no parentheses, that "$\ln 2^2$" really means "$\ln (2^2)$". If you intend "$(\ln 2)^2$", you should explicitly write paretheses. Likewise with expressions like "$\sin x^2$", which would mean "$\sin(x^2)$".

MPW
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I am left with $\ln(2)^2$ so I move the power down using the log rule and do $16\cdot 2$.

You seem to assume $(\ln 2)^2 = 2 \ln(2)$ [wrong] instead of $\ln(2^2) = 2 \ln(2)$ [correct, but not the case here] and $(\ln(2))^2 = \ln(2) \ln(2) = \ln(2^{\ln(2)})$ [correct, but not really easier].

So I would stay at $(\ln(2))^2$ as well.

mvw
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