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Given this equation

$$5^m - 7^m = -2$$ It is clear that the $m = 1$ but how do we solve it?

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    what is $m$ here? – 2017-01-25
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    @Dr.SonnhardGraubner the unknown. – 2017-01-25
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    yes i know, but what Kind of unknown? – 2017-01-25
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    In which set of numbers does $m$ come? Dr. S Graubner meant to say that. – 2017-01-25
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    It's Real numbers. – 2017-01-25
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    It'll be better to add this back into the question body, so that others don't need to read the comments in order to know $m \in \Bbb R$. – 2017-01-25

3 Answers 3

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If $m>1$ then $$ 7^m-5^m=7^m\Bigl(1-\frac{5^m}{7^m}\Bigr)>7^m\Bigl(1-\frac{5}{7}\Bigr)=2\cdot7^{m-1}>2. $$ You can argue similarly if $m<1$. This leaves $m=1$ as the only solution.

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    What is the similar argument for $0 – 2017-01-25
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    If $m<0$ then $5^m-7^m<5^m<1$, and it can never equal $-2$. This leave $0\le m<1$. $$7^m-5^m=7^m\Bigl(1-\frac{5^m}{7^m}\Bigr)<\dots$$ You can finish. – 2017-01-25
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Let $f(x)=7^x-5^x$. We want to prove that $f(x)=2$ has a unique solution. Here is a sketch of the proof:

The derivative of $f$ is $$f'(x)=\ln 7 \cdot 7^x - \ln 5 \cdot 5^x$$ and $f'(x) > 0$ is equivalent to $$x > \frac{\ln \log_7 5}{\ln 7/5} = A$$

$f$ is decreasing on $(- \infty , A)$, and $\lim_{x \to -\infty} f(x)=0$: these two information give us $f(x)<0$ for $x \in (- \infty , A]$, so that any solution must be $>A$. But $f$ is increasing on $(A , + \infty)$ , so that any solution must be unique.

The range of $f$ is indeed $[f(A) , + \infty)$ , since $f(x) \to \infty$ as $x \to \infty$ and $f$ is continuous. Since $2>0=f(0) > f(A)$ we have that $$f(x)=2$$ has a solution, and it is unique by what we said above.

Once you notice that $x=1$ is your solution, you are done.

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$$\log_{5}x=m$$

$$5^m - 7^m = -2$$

$$a^{\log(b)}=b^{\log(a)}$$

$$5^{\log_{5}x} - 7^{\log_{5}x} = -2$$

$${x} - x^{\log_{5}7} = -2$$

$${x} +2 = x^{\log_{5}7}$$

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just $x=5$ => $m=1$