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$ 2 \ln (5x) = 16$

$ \ln (5x) = 8 $

$ 5x = e^8 $

$ x = \dfrac {1}{5}e^8$

But why can't we do it like this:

$ \ln(5x)^2 = 16$

I thought that was a possibilty with logaritms?

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$2 \ln{5x}= \ln{(25x^2)} \neq \ln(5x)^2$ –  Bob Jan 21 '13 at 22:13
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@Bob, CoffeeIsStupid: The notation is ambiguous. I would add extra parentheses somewhere to clarify: $\ln((5x)^2)$ or $(\ln(5x))^2$? The first is correct if you add the restriction $x>0$ to maintain equivalence (because $\ln(5x)$ is only defined when $x>0$), but the second is incorrect. –  Jonas Meyer Jan 21 '13 at 22:18
    
@JonasMeyer: by convention I consider the notation $\ln{(5x)}^2$ to be used wrong here. (since double brackets would be overkill) –  Bob Jan 21 '13 at 22:22
    
@Bob: Your convention disagrees with that of some others; that is why it is ambiguous. –  Jonas Meyer Jan 21 '13 at 22:25
    
@Bob I suspect you mean to say $\ln(25x^2) \not\equiv \ln(5x)^2$. For there are certainly values for which $\ln(25x^2) = \ln(5x)^2$, for example $x=\frac{1}{5}$ or $x=\frac{1}{5}e^2$. –  Fly by Night Jan 21 '13 at 22:37
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3 Answers

Sure it's possible. We have $\ln((5x)^2)=16$, and therefore $(5x)^2=e^{16}$. Take square roots, remembering that $x$ must be positive for the logarithm to be defined. We get the right answer, a little more slowly than before.

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$$\ln ((5x)^2) = 16,\qquad x>0$$

$$25x^2 = e^{16}$$

$$x^2 = \frac{e^{16}}{25}$$

$$x = \pm\sqrt{\frac{e^{16}}{25}} =\pm\frac{\sqrt{e^{16}}}{5}=\pm\frac{e^{16/2}}{5}=\pm \frac{e^8}{5}$$

But since you've introduced the square you have to go back and check the answers - The negative one doesn't fit. So they're equivalent.

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If you mean $\ln\left[(5x)^2\right]$, then yes!

You are allowed to do that!

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But be careful to note that the restriction $x>0$ must be added to maintain equivalence to the original equation. –  Jonas Meyer Jan 21 '13 at 22:16
    
Sure, but that's not exactly as was asked. With the brackets it would be ok. –  Bob Jan 21 '13 at 22:16
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