# Why does $7^{2\ln x}\cdot \ln(7) \cdot (2/x)$ equal to $7^{2\ln x}\cdot \ln(49) /x$?

While reviewing, I came upon this problem which has the derivative

$7^{2\ln x}\cdot \ln(7) \cdot (2/x)$

simplified to

$7^{2\ln x}\cdot \ln(49) /x$

How/why is it simplified like that?

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## 2 Answers

The simplification relied on the following fact:

If $a>0$, then $2\log(a) = \log(a^2)$.

In your case, the $2 \log(7)$ got converted to $\log(7^2) = \log(49)$. The rest of the terms remain as such.

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That was fast. Thank you! – Gannicus May 15 '13 at 5:56

You can use the rules of exponents and logarithms to come up with a few equivalent forms. For example

$$7^{2\ln x}=(7^2)^{\ln x}=49^{\ln x}$$

If you take the derivative of that, you'll get the form you seek. It's also possible to rewrite the original as

$$7^{2\ln x}=49^{\ln x}=(e^{\ln 49})^{\ln x}=(e^{\ln x})^{\ln 49}=x^{\ln 49}$$

Take the derivative of that to get

$$x^{\ln49-1}\ln49=x^{\ln49-\ln e}\ln 49=x^{\ln\frac{49}e}\ln49$$

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