Solve for x: e^x - 15e^-x = 2 | Algebraic Solution & Step-by-Step Guide

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im having a problem solving this problem algebraically please help thanks

e^x - 15e^-x = 2
 
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put e^x = y. And then solve for y
 
Remember the rule: a.b^{-c}=\frac{a}{b^c}

You will have a quadratic in e^x
 

Thread 'Greatest possible value of a constant in polynomial'

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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