Question

4^x+6^x=9^x
please solve this​

Answer 1

Answer:

Step-by-step explanation:

$4^x+6^x=9^x\\\\Divide\;\;by \;\;4^x\\\\let \;\;(3/2)^x \;\;=y\\\\then\\\\y^2-2y-1=0\\\\y^2-2y+1=2\\\\(y-1)^2=2\\\\y-1= {\sqrt{2}}\\\\y= \sqrt{2} +1\\\\y-1=-\sqrt{2}\\\\y=1-\sqrt{2}\\\\we \;\;have\;\;y=(3/2)^x\\\\x=log_{3/2}y$$Since\;\;log\;\;of\;\;negative \;numbers\;is\;not\;valid\\\\y \neq 1-\sqrt{2}\\\\=> y=\sqrt{2}+1\\\\x=log_{3/2}(\sqrt{2}+1)$

Hope it Helps

Answer 2

Answer:

my Facebook name is :

virendra ramina

Step-by-step explanation:

my instagram name is :

virendraramina1822

my twitter name is

virendraramina18

plzzz follow on all of this 0lzzz and mark as brainlist