Exponential Function: $4^x + 6^x = 9^x$
Let $x$ be a real number that satisfy the equation
$4^x + 6^x = 9^x$
What is the value of $\left(\dfrac{2}{3}\right)^x$ ?
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Dividing both sides by 6x,
Dividing both sides by 6x, you get
4x/6x+1=9x/6x
Simplifying the fractions you get
2x/3x+1=3x/2x
That can be written as
(2/3)x+1=(3/2)x
Now, if the number you want is (2/3)x=k, then (3/2)x=1/k ,
which means the equation can be written as
k+1=1/k
Multiplying both sides times k,
k2+k=1
Solving that quadratic equation gives you two real solutions for k.
There is something misleading
In reply to Dividing both sides by 6x, by KMST
There is something misleading hahaha.
Do the two real solutions for k BOTH give REAL solutions for $x$ ?
I am not used anymore to
I am not used anymore to calculate this type of equation, the result of relying too much in
SHIFT + SOLVEof Casio. I did not say doingShift + Solveis bad, it is actually highly recommended in my line of work, hehehe. Anyway, allow me to solve this, and I am actually surprised that solving for x is more complex than solving for (2/3)x. Here is my take based on the suggestion of KMST.$4^x + 6^x = 9^x$
$\dfrac{4^x}{6^x} + \dfrac{6^x}{6^x} = \dfrac{9^x}{6^x}$
$\left( \dfrac{4}{6} \right)^x + 1 = \left( \dfrac{9}{6} \right)^x$
$\left( \dfrac{2}{3} \right)^x + 1 = \left( \dfrac{3}{2} \right)^x$
$\left( \dfrac{2}{3} \right)^x + 1 = \dfrac{1}{\left( \dfrac{2}{3} \right)^x}$
$\left( \dfrac{2}{3} \right)^{2x} + \left( \dfrac{2}{3} \right)^x = 1$
$\left( \dfrac{2}{3} \right)^{2x} + \left( \dfrac{2}{3} \right)^x - 1 = 0$
By Quadratic Equation
$\left( \dfrac{2}{3} \right)^x = \dfrac{-1 \pm \sqrt{1^2 - 4(1)(-1)}}{2(1)}$
$\left( \dfrac{2}{3} \right)^x = \dfrac{-1 \pm \sqrt{5}}{2}$
$\left( \dfrac{2}{3} \right)^x = \dfrac{-1 + \sqrt{5}}{2}$
$\log \left( \dfrac{2}{3} \right)^x = \log \dfrac{-1 + \sqrt{5}}{2}$
$x \log \left( \dfrac{2}{3} \right) = \log (\sqrt{5} - 1) - \log 2$
$x (\log 2 - \log 3) = \log (\sqrt{5} - 1) - \log 2$
$x = \dfrac{\log (\sqrt{5} - 1) - \log 2}{\log 2 - \log 3}$ ← a real number
For
$\left( \dfrac{2}{3} \right)^x = \dfrac{-1 - \sqrt{5}}{2}$
$\log \left( \dfrac{2}{3} \right)^x = \log \dfrac{-1 - \sqrt{5}}{2}$
$x \log \left( \dfrac{2}{3} \right) = \log (-\sqrt{5} - 1) - \log 2$
$x (\log 2 - \log 3) = \log (-\sqrt{5} - 1) - \log 2$
$x = \dfrac{\log (-\sqrt{5} - 1) - \log 2}{\log 2 - \log 3}$ ← underfined
Hence,
$\left( \dfrac{2}{3} \right)^x = \dfrac{\sqrt{5} - 1}{2}$ ← this is my answer.
Please be gentle with me, hehehe. Although I hope I was able to consider everyhting, maybe I miss simething. As I've said, I am no longer used to this type of approach. A decimal number from calculator is more than acceptable for me.