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Edit: I just rewrote the whole question, to make it clear, what I'm looking for. Originally I asked about solving $16^{x+1} = 4^{x+3}$, then corrected it to $16^{x+2} = 4^{x+3}$.

But what I really want to know is this: How do you get from

$(x + 2)\ln(16) - (x + 3)\ln(4) = \ln(1)$

to

$x = \frac{3\ln(4) - 2\ln(16)}{\ln(16) - \ln(4)}$

I'm particularly concerned about the factors $(x+2)$ and $(x+3)$ as I don't understand, how they lead up the the last term above.

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    I'm not particularly advocating standards, just trying to make the whole page make sense when you visit it the first time :)2012-05-17

3 Answers 3

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Once you've got $(x+2)\ln(16)−(x+3)\ln(4)=\ln(1)$, use $\log 1 = 0$ (true regardless of the base) and then forget about the logs entirely for now: the rest is just algebra, so let $a = \ln(16)$ and $b = \ln(4)$. Then solve $(x+2)a - (x+3)b = 0$ for $x$, by expanding the brackets and rearranging; you will get $x = \frac{3b - 2a}{a - b}$, which is indeed (substituting back in $a$ and $b$) exactly what you wanted.

(If you can't do the algebra, say so and I'll give a bit more detail).

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    Ah, now it makes sense. I got too focused on the logs and somehow fumbled over the whole equation.2012-05-17
4

To proceed with what you have done, you need to recognize that $\ln(1) = 0$ and $\ln(a^b) = b \times \ln(a)$ i.e. $\ln(16) = \ln(4^2) = 2 \ln(4)$. Then you can simplify the expression you have to get $(x+2) \times 2 \times \ln(4) - (x+3) \times \ln(4) = 0$ i.e. \begin{align*} (x+2) \times 2 \times \ln(4) &= (x+3) \times \ln(4) \\ 2(x+2) &= x+3 \\ 2x+4 &= x +3 \\ x&=-1 \end{align*}

The easier way is to take the logarithm to the base $4$. Then we get that $(x+2) \times \log_4(16) = (x+3) \times \log_4(4)$ Hence, \begin{align*} (x+2) \times \log_4(4^2) &= (x+3) \times 1 \\ (x+2) \times 2 \times \log_4(4) &= (x+3) \times 1 \\ (x+2) \times 2 \times 1 &= (x+3) \times 1 \\ 2x+4 &= x+3 \\ x&=-1 \end{align*}

4

Why would you apply logarithms before simplifying stuff??

$16^{x+1}=4^{x+3}\Longrightarrow \left((2^4)\right)^{x+1}=\left(2^2\right)^{x+3}\Longrightarrow 2^{4x+4}=2^{2x+6}\Longrightarrow 4x+4=2x+6\Longrightarrow x=1$

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    Of course it is wrong! $,1=16^0=16^{-1+1}\neq 4^{-1+2}=4^2=16$2012-05-17