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4^x + 6^x = 9^x

find the value of X.

please help me with this!!!

please

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Question

4^x + 6^x = 9^x

find the value of X.

please help me with this!!!

please

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Mathematics
2 years
2021-07-15T14:48:49+00:00
2021-07-15T14:48:49+00:00 2 Answers
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## Answers ( )

Given:ㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤORweHaveㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤORStep-by-step explanation:Given:–4^x + 6^x = 9^x

Tofind:–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

Usedformulae:–