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# Challenge problem!

0
193
5

$$4^x+6^x=9^x$$

Solve for x

Nov 22, 2019

### 5+0 Answers

#1
0

x≈1.18681

Nov 22, 2019
#2
+1

yeah correct, showing the steps is challenging tho.

Guest Nov 22, 2019
#3
+109518
+2

4^x+6^x=9^x

$$4^x+6^x=9^x\\ 2^x(2^x+3^2)=3^x3^x\\ 2^x+3^x=3^x\cdot (\frac{3}{2})^x\\ (\frac{2}{3})^x+1= (\frac{3}{2})^x\\ let\;\;y=2/3\\ y^x+1=(\frac{1}{y})^x\\ y^{2x}+y^x=1\\ (y^{x})^2+y^x-1=0\\ y^x=\frac{-1\pm\sqrt{1+4}}{2}\\ y^x \;\;must\;\;be \;\;pos\;\;so\;\;\\ y^x=\frac{-1+\sqrt{5}}{2}\\ log(y^x)=log(\frac{-1+\sqrt{5}}{2})\\ xlog(y)=log(\frac{-1+\sqrt{5}}{2})\\ xlog(\frac{2}{3})=log(\frac{-1+\sqrt{5}}{2})\\ x=log(\frac{-1+\sqrt{5}}{2})\div log(\frac{2}{3})\\ x\approx 1.1868$$

Latex coding

4^x+6^x=9^x\\
2^x(2^x+3^2)=3^x3^x\\
2^x+3^x=3^x\cdot (\frac{3}{2})^x\\
(\frac{2}{3})^x+1= (\frac{3}{2})^x\\
let\;\;y=2/3\\
y^x+1=(\frac{1}{y})^x\\
y^{2x}+y^x=1\\
(y^{x})^2+y^x-1=0\\
y^x=\frac{-1\pm\sqrt{1+4}}{2}\\
y^x \;\;must\;\;be \;\;pos\;\;so\;\;\\
y^x=\frac{-1+\sqrt{5}}{2}\\
log(y^x)=log(\frac{-1+\sqrt{5}}{2})\\
xlog(y)=log(\frac{-1+\sqrt{5}}{2})\\
xlog(\frac{2}{3})=log(\frac{-1+\sqrt{5}}{2})\\
x=log(\frac{-1+\sqrt{5}}{2})\div log(\frac{2}{3})\\
x\approx 1.1868

Nov 22, 2019
#4
+1

Amazing solution Melody!!

Here is the solution :

https://www.youtube.com/watch?v=6AwfRXKvGsM

Guest Nov 22, 2019
#5
+109518
0

Thanks,

The video solution is exactly the same as mine.

( I did solve it by myself though)

Melody  Nov 22, 2019