Exponential Function: 4^x + 6^x = 9^x
New forum topics
- Please help me solve this problem: Moment capacity of a rectangular timber beam
- Solid Mensuration: Prismatoid
- Differential Equation: (1-xy)^-2 dx + [y^2 + x^2 (1-xy)^-2] dy = 0
- Differential Equation: y' = x^3 - 2xy, where y(1)=1 and y' = 2(2x-y) that passes through (0,1)
- Tapered Beam
- Vickers hardness: Distance between indentations
- Time rates: Question for Problem #12
- Make the curve y=ax³+bx²+cx+d have a critical point at (0,-2) and also be a tangent to the line 3x+y+3=0 at (-1,0).
- Minima maxima: Arbitrary constants for a cubic
- Minima Maxima: y=ax³+bx²+cx+d
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
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.