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$$9^m + 9^n = 52$$ $$9^m -4 = 2 \cdot 9^n$$

$$3^{m-n}=?$$

Let me show what I've tried

Simpifyling the both equalities.

$$3^{2^m} + 3^{2n} = 2 \cdot 13 \tag{1}$$ $$3^{2m} -2^2 = 2 \cdot 3^{2n} \tag{2}$$

Diving the second equality by $2$ and we have

$$\frac{3^{2m} -2^2}{2} =3^{2n} \tag{3}$$

Here is where I'm stuck.

My Kindest Regards!

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    $\begingroup$ Well, write your equations as $x+y=52,\;x-2y=4$. Solve for $x,y$. $\endgroup$ – lulu Apr 25 '18 at 18:03
  • $\begingroup$ There are no solutions for integers $n,m\ge 1$. Usually $9^x+9^y=52$ is written for real $x,y$. $\endgroup$ – Dietrich Burde Apr 25 '18 at 18:06
  • $\begingroup$ Normally problems like this are set up for $n,m$ integers, but that is not guaranteed. Here the solution is not integral. Did you copy it correctly? $\endgroup$ – Ross Millikan Apr 25 '18 at 18:10
  • $\begingroup$ @RossMillikan Yes, I did. $\endgroup$ – Fiv Apr 25 '18 at 18:21
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Putting $$ x=9^{m}, \quad y=9^{n}, $$ we have

\begin{eqnarray*} x+y&=&52\\ x-2y&=&4 \end{eqnarray*} It follows that $$ 9^{m}=x=\frac{3x}{3}=\frac{108}{3}=36, \quad 9^{n}=y=\frac{3y}{3}=\frac{48}{3}=16, $$ i.e. $$ 3^{m}=\sqrt{9^{m}}=\sqrt{36}=6,\quad 3^{n}=\sqrt{9^{n}}=\sqrt{16}=4. $$ Hence $$ 3^{m-n}=\frac{3^{m}}{3^{n}}=\frac{6}{4}=\frac{3}{2}. $$

Remark: The numbers $m$ and $n$ are obviously not integers, in fact $$ m=\log_3(6)=\frac{\ln(6)}{\ln(3)}\approx 1.63092975\ldots ,\quad n=\log_3(4)=\frac{\ln(4)}{\ln(3)}\approx 1.261859507\ldots $$

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  • $\begingroup$ The answer is given as $\frac{3}{2}$. $\endgroup$ – Fiv Apr 25 '18 at 18:22
  • $\begingroup$ @Fiv I don't understand your comment b/c the answer is given before the remark. $\endgroup$ – Mercy King Apr 25 '18 at 18:24
  • $\begingroup$ Oh I didn't see that lol! and I didn't understand it. $\endgroup$ – Fiv Apr 25 '18 at 18:26
  • $\begingroup$ I'd suggest you Fiv , to learn some basic maths before coming on SE $\endgroup$ – Ravi Prakash Apr 27 '18 at 4:09
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$$9^m + 9^n = 52$$ $$9^m -4 = 2 \cdot 9^n$$

By substituion, we have

$$(52-9^n)-4=2\cdot 9^n$$

$$48-9^n = 2 \cdot 9^n$$

$$9^n = 16$$

$$3^n=4$$

Substitute $9^n$ inside the first equation and do the same trick to complete the task.

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  • $\begingroup$ I didn't get this step: $(52-9^n)-4=2\cdot 9^n$ $\endgroup$ – Fiv Apr 25 '18 at 18:25
  • $\begingroup$ from the first $9^m =52-9^n$, substitute that into the second equation. $\endgroup$ – Siong Thye Goh Apr 25 '18 at 18:40

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