Solve 7^2x+3=7^x^2: Step-by-Step Guide

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AI Thread Summary
The equation 7^(2x+3) / 7^(x^2) = 1 can be solved by cross-multiplying or rewriting the terms for clarity. It's important to express the exponent correctly, using 7^(x^2) to avoid confusion. Participants discussed the correct interpretation of the numerator, confirming it should be 7^(2x + 3). After some guidance, one user successfully solved the equation. Clear notation is essential for solving exponential equations effectively.
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Homework Statement


Solve the equation
7^2x+3 / 7^x^2 = 1

Homework Equations





The Attempt at a Solution



How can i "breakdown" 7^x^2? Thank you!
 
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Are you sure breaking down 7^x^2 is the only way?
Try cross multiplying it to the other side and observe.
 
AlchemistK said:
Are you sure breaking down 7^x^2 is the only way?
Try cross multiplying it to the other side and observe.

Oh I got it! I managed to solve it =) thanks!
 
Mphisto said:

Homework Statement


Solve the equation
7^2x+3 / 7^x^2 = 1
Is the numerator supposed to be 72x + 3 or 72x + 3 or 72x + 3?

I suspect that it's the first. If that's what you meant, write it as 7^(2x + 3) so that it is unambiguous.
Mphisto said:

Homework Equations





The Attempt at a Solution



How can i "breakdown" 7^x^2? Thank you!
 
Mphisto said:
Solve the equation
7^2x+3 / 7^x^2 = 1

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How can i "breakdown" 7^x^2? Thank you!

And, to be unambiguous, type "7^(x^2)"

if you mean 7^{x^2}.
 
checkitagain said:
And, to be unambiguous, type "7^(x^2)"

if you mean 7^{x^2}.

Yeah, I meant that
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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