Logarithmic Question: Let e^(2x) - ae^x + 1/6 = 0, w/ sol'n log_q^2(p), log_p^3(q), p

[flyingcloud]

Let e^2x-ae^x+1/6=0, and this equation has the solution of log_q² (p) and log_p³ (q).
Given that p>1, q>1. The minimum value of a happens when log_p (q) = y. Find the value of a and y.

I cannot solve this. Please help me.

 

[stapel]

Let e^2x-ae^x+1/6=0, and this equation has the solution of log_q² (p) and log_p³ (q).
Given that p>1, q>1. The minimum value of a happens when log_p (q) = y. Find the value of a and y.

I cannot solve this.

How does your book approach similar exercises? Given that you're in algebra, not calculus, what methods are you supposed to be using for finding minimization values? What have you tried? Where are you stuck?

For instance, you know that \(\displaystyle \,e^{2x}\, -\, a\, e^x\, +\, {}^1/_6\, =\, 0\,\) is a quadratic in \(\displaystyle \,e^x\,\) so you applied the Quadratic Formula and... then what?

Please be complete. Thank you!