Question

4^x + 6^x = 9^x

find the value of X.




please help me with this!!!

please

​

Answer 1

$\huge \fbox \pink{Solutíon:}$

Given:

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

ㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤOR

$( \frac{4}{9} {)}^{x} + ( \frac{2}{3} {)}^{x} = 1$

$Putting \: (2/3) = y$

we Have

${y}^{2} + y - 1 = 0$

ㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤOR

$y = \frac{ - 1≠ \sqrt{5} }{2}$

$⇒( \frac{2}{3} {)}^{x} = \frac{ \sqrt{5} - 1}{2}$

$⇒x = log_{2/3}( \frac{ \sqrt{5} - 1}{2} )$

$\\ \\ \\ \sf \colorbox{gold} {\red(ANSWER ᵇʸ ⁿᵃʷᵃᵇ⁰⁰⁰⁸}$

Answer 2

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