"Easy" exponential eqn: 3^x+3^{x+1}+3^{x+2} = 5^x+5^{x+1}+5^{x+2}

granitba

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So we held this short quiz/test and all the questions were really easy, the equation below seemed like I'd be able to solve it in under a minute but I spent 30min and couldn't solve it.

3x+3x+1+3x+2=5x+5x+1+5x+2

All I can do is factor 3x on the left and 5x​ on the right.
 
So we held this short quiz/test and all the questions were really easy, the equation below seemed like I'd be able to solve it in under a minute but I spent 30min and couldn't solve it.

3x+3x+1+3x+2=5x+5x+1+5x+2

All I can do is factor 3x on the left and 5x​ on the right.
Please share your work - we will show the next step accordingly.
 
So we held this short quiz/test and all the questions were really easy, the equation below seemed like I'd be able to solve it in under a minute but I spent 30min and couldn't solve it.

3x+3x+1+3x+2=5x+5x+1+5x+2

All I can do is factor 3x on the left and 5x​ on the right.
You started well with the factoring. Why stop?

\(\displaystyle 3^x + 3^{(x + 1)} + 3^{(x+2)} = 5^x + 5^{(x+1)} + 5^{(x+2)} \implies\)

\(\displaystyle 3^x(1 + 3 + 9) = 5^x(1 + 5 + 25) \implies 13 * 3^x = 31 * 5^x \implies\)

\(\displaystyle \dfrac{31}{13} = \dfrac{3^x}{5^x} = \left ( \dfrac{3}{5} \right )^x \implies WHAT?\)
 
You started well with the factoring. Why stop?

\(\displaystyle 3^x + 3^{(x + 1)} + 3^{(x+2)} = 5^x + 5^{(x+1)} + 5^{(x+2)} \implies\)

\(\displaystyle 3^x(1 + 3 + 9) = 5^x(1 + 5 + 25) \implies 13 * 3^x = 31 * 5^x \implies\)

\(\displaystyle \dfrac{31}{13} = \dfrac{3^x}{5^x} = \left ( \dfrac{3}{5} \right )^x \implies WHAT?\)


That's how far I got too, but I have no idea what I'm supposed to do from there on.
 
You started well with the factoring. Why stop?

\(\displaystyle 3^x + 3^{(x + 1)} + 3^{(x+2)} = 5^x + 5^{(x+1)} + 5^{(x+2)} \implies\)

\(\displaystyle 3^x(1 + 3 + 9) = 5^x(1 + 5 + 25) \implies 13 * 3^x = 31 * 5^x \implies\)

\(\displaystyle \dfrac{31}{13} = \dfrac{3^x}{5^x} = \left ( \dfrac{3}{5} \right )^x \implies WHAT?\)

I already got that far, I have no idea what to do next, and I'm certain it doesn't involve logarithms.
 
I already got that far
Okay; in future, please don't claim to have had no idea how to get past an earlier step, if you have indeed gotten further. In all cases, please show your work.

I have no idea what to do next, and I'm certain it doesn't involve logarithms.
On what basis have you concluded that the solution method does not involve taking logs?

Please be complete. Thank you! ;)
 
On the fact that the other groups had almost the same equation but had a whole number as an answer.
At what point did their work diverge? On what basis have you concluded that your different equation should have the same solution?

Please be complete, including showing their equation. Thank you! ;)
 
I already got that far, I have no idea what to do next, and I'm certain it doesn't involve logarithms.
Consider any equation of the form

\(\displaystyle a = b^z \text {, where } a > 0 < b \text { and } b \ne 1.\)

\(\displaystyle \therefore log(a) = log(b^z) = z * log(b) \implies z = \dfrac{log(a)}{log(b)}\)

How does that general logic apply to your problem?
 
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