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

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

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

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!

• Well, write your equations as $x+y=52,\;x-2y=4$. Solve for $x,y$. – lulu Apr 25 '18 at 18:03
• There are no solutions for integers $n,m\ge 1$. Usually $9^x+9^y=52$ is written for real $x,y$. – Dietrich Burde Apr 25 '18 at 18:06
• 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? – Ross Millikan Apr 25 '18 at 18:10
• @RossMillikan Yes, I did. – Fiv Apr 25 '18 at 18:21

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 $$m=\log_3(6)=\frac{\ln(6)}{\ln(3)}\approx 1.63092975\ldots ,\quad n=\log_3(4)=\frac{\ln(4)}{\ln(3)}\approx 1.261859507\ldots$$

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

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

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.

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