[tex]\huge \fbox \pink{Solutíon:}[/tex] Given: [tex] {4}^{x} + {6}^{x} = {9}^{x} [/tex] ㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤOR [tex]( \frac{4}{9} {)}^{x} + ( \frac{2}{3} {)}^{x} = 1[/tex] [tex]Putting \: (2/3) = y[/tex] we Have [tex] {y}^{2} + y – 1 = 0[/tex] ㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤOR [tex]y = \frac{ – 1≠ \sqrt{5} }{2} [/tex] [tex]⇒( \frac{2}{3} {)}^{x} = \frac{ \sqrt{5} – 1}{2} [/tex] [tex]⇒x = log_{2/3}( \frac{ \sqrt{5} – 1}{2} )[/tex] [tex] \\ \\ \\ \sf \colorbox{gold} {\red(ANSWER ᵇʸ ⁿᵃʷᵃᵇ⁰⁰⁰⁸}[/tex] Reply
Step-by-step explanation: Given:– 4^x + 6^x = 9^x To find :– Find the value of x ? Solution:– Given equation is 4^x + 6^x = 9^x On dividing by 9^x both sides then =>(4^x / 9^x )+ (6^x /9^x) = 9^x /9^x =>(4^x / 9^x )+ (6^x /9^x) = 1 We know that a^m/b^m = (a/b)^m =>(4/9)^x +(6/9)^x = 1 =>(4/9)^x + (2/3)^x = 1 =>[(2/3)^2]^x +(2/3)^x = 1 We know that (a^m)^n = a^mn =>(2/3)^2x +(2/3)^x = 1 Put (2/3)^x = a then =>a^2 + a = 1 =>a^2 +2(a)(1/2) = 1 On adding (1/2)^2 both sides =>a^2+2(a)(1/2)+(1/2)^2 = 1+(1/2)^2 =>[a+(1/2)]^2 = 1+(1/4) =>[a+(1/2)]^2 = (4+1)/4 =>[a+(1/2)]^2 = 5/4 =>a +(1/2) = ±√(5/4) =>a +(1/2) = ±√5/2 =>a = ±√5/2 -(1/2) =>a = ±(√5-1)/2 Since x is a positive then the square of a positive number is always a positive number =>a = (√5-1)/2 now , (2/3)^x = (√5-1)/2 On applying logarithmic form We know that a^x = N => log N (a) = x Wher (a) is base a =>x = log (√5-1)/2 (2/3) Here base 2/3) And it also can be written as =>x = log (√5-1)-log2 (2/3) (log (√5-1)-log 2 to the base 2/3) Answer:– answer for the given problem is x = log (√5-1)/2 (2/3) where 2/3 is a base Used formulae:– a^m/b^m = (a/b)^m (a^m)^n = a^mn a^x = N => log N (a) = x Where a is a base Reply
[tex]\huge \fbox \pink{Solutíon:}[/tex]
Given:
[tex] {4}^{x} + {6}^{x} = {9}^{x} [/tex]
ㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤOR
[tex]( \frac{4}{9} {)}^{x} + ( \frac{2}{3} {)}^{x} = 1[/tex]
[tex]Putting \: (2/3) = y[/tex]
we Have
[tex] {y}^{2} + y – 1 = 0[/tex]
ㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤOR
[tex]y = \frac{ – 1≠ \sqrt{5} }{2} [/tex]
[tex]⇒( \frac{2}{3} {)}^{x} = \frac{ \sqrt{5} – 1}{2} [/tex]
[tex]⇒x = log_{2/3}( \frac{ \sqrt{5} – 1}{2} )[/tex]
[tex] \\ \\ \\ \sf \colorbox{gold} {\red(ANSWER ᵇʸ ⁿᵃʷᵃᵇ⁰⁰⁰⁸}[/tex]
Step-by-step explanation:
Given:–
4^x + 6^x = 9^x
To find :–
Find the value of x ?
Solution:–
Given equation is 4^x + 6^x = 9^x
On dividing by 9^x both sides then
=>(4^x / 9^x )+ (6^x /9^x) = 9^x /9^x
=>(4^x / 9^x )+ (6^x /9^x) = 1
We know that a^m/b^m = (a/b)^m
=>(4/9)^x +(6/9)^x = 1
=>(4/9)^x + (2/3)^x = 1
=>[(2/3)^2]^x +(2/3)^x = 1
We know that (a^m)^n = a^mn
=>(2/3)^2x +(2/3)^x = 1
Put (2/3)^x = a then
=>a^2 + a = 1
=>a^2 +2(a)(1/2) = 1
On adding (1/2)^2 both sides
=>a^2+2(a)(1/2)+(1/2)^2 = 1+(1/2)^2
=>[a+(1/2)]^2 = 1+(1/4)
=>[a+(1/2)]^2 = (4+1)/4
=>[a+(1/2)]^2 = 5/4
=>a +(1/2) = ±√(5/4)
=>a +(1/2) = ±√5/2
=>a = ±√5/2 -(1/2)
=>a = ±(√5-1)/2
Since x is a positive then the square of a positive number is always a positive number
=>a = (√5-1)/2
now ,
(2/3)^x = (√5-1)/2
On applying logarithmic form
We know that a^x = N => log N (a) = x
Wher (a) is base a
=>x = log (√5-1)/2 (2/3)
Here base 2/3)
And it also can be written as
=>x = log (√5-1)-log2 (2/3)
(log (√5-1)-log 2 to the base 2/3)
Answer:–
answer for the given problem is
x = log (√5-1)/2 (2/3)
where 2/3 is a base
Used formulae:–