How would I solve for $x$ in $4^{x+7}=5^{x-3}$?

Question

So I was bored, and decided to see if I was still able to solve logarithmic equations when I came up with the question$$\text{Solve for }x\text{: }4^{x+7}=5^{x-3}$$which I thought that I might be able to solve. Here is my attempt at solving the aforementioned equation:$$4^{x+7}=5^{x-3}$$$$4^{x+7}=4^{x-3}\cdot1.25^{x-3}$$$$4^{\require{cancel}\cancel{x}+7\cancel{-x}+3}=\dfrac{\cancel{4^{x-3}}\cdot1.25^{x-3}}{\cancel{4^{x-3}}}\qquad(\because a^b\cdot c^b=d^b,ac=d)$$$$4^{10}=1.25^{x-3}$$$$10\ln4=(x-3)\ln1.25$$$$\dfrac{10\ln4}{x-3}=\ln1.25$$$$\dfrac{x-3}{10\ln4}=\dfrac1{\ln1.25}$$$$x-3=\dfrac{10\ln4}{\ln1.25}$$$$x=\dfrac{10\ln4}{\ln1.25}+3$$$$\therefore\text{ for any }a^{x+b}=c^{x+d},a\neq c,\text{ }a,b,c,d\in\Bbb{R},\text{ }a,c\gt0$$$$x=\dfrac{(b-d)\ln(a)}{\ln\dfrac ca}-d$$

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My question

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Is the solution that I have achieved correct, or what could I do to attain the correct solution more easily/quickly?

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Mistakes that I might have made

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1. Simplifying the logarithms (if you read a lot of my posts you might understand why this is a problem)
2. Using reciprocals (basically raising both sides to the $-1$st power)
3. Incorrect tags on this question

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For anyone who doesn't understand why something is the way I am defining it

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"$\dots,a\neq c\dots a,c\gt0$"

This is because if $a=0$, not only can you not take a logarithm of $0$, you also cannot divide by $0$. If $c=0$, you cannot take a logarithm of $0$. (since it is undefined) If $a=c$, this is fine until you realize any $\dfrac ab$, $a=b$ is equal to $1$, and even though the logarithm of that is defined in any base (that value being $0$), you cannot divide by $0$. However, this is not to say that this will never be true if $a=c=1$, since then it will be all solutions. However, if $|a=c|\gt1$, then only if $b=d$ will there be any solutions, albeit an infinite amount, however, if $|a=c|\gt1,b\neq d$, there will be no solutions since it will never equal each other anywhere.

Answer 1

Here's the step-by-step solution:

Take the natural logarithm (ln) of both sides:

$\ln(4^{x+7})=\ln(5^{x-3})$

Apply the logarithmic property:

$\ln(a^b) = b\ln(a) \cdot(x+7)\ln(4)=(x-3)\ln(5)$

Distribute the $\ln(4)$ and $\ln(5)$ across the terms:

$x\ln(4)+7\ln(4)=x\ln(5)-3\ln(5)$

Move all terms involving x to one side and constants to the other side:

$x\ln(4)-x\ln(5)=-3\ln(5)-7\ln(4)$

Factor out $x$:

$x(\ln(4)-\ln(5))=-3\ln(5)-7\ln(4)$

Divide both sides by $\ln(4)-\ln(5)$:

$x=\dfrac{-3\ln(5)-7\ln(4)}{\ln(4)-\ln(5)}$

Answer 2

You could just take log of both sides directly. For $a^{x+b} = c^{x+d}$, $$ (x+b) \log(a) = (x+d) \log(c)$$ and then solve for $x$: $$ x = \frac{d \log(c) - b \log(a)}{\log(a)- \log(c)}$$

Answer 3

This looks fine, but it may be considered simpler to just do \begin{align} & (x+7) \ln 4 = (x-3) \ln 5 \\ & 7 \ln 4 + 3 \ln 5 = x \ln 5/4 \\ & x = \frac{7 \ln 4 + 3 \ln 5}{\ln5/4} \end{align}