Question

After solving this you will get a golden answer.
find the value of x
${4}^{x} + {6}^{x} = {9}^{x}$

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Answer 1

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 4^x both sides , then

=> (4^x+6^x)/(4^x) = (9^x)/(4^x)

=> (4^x/4^x) +(6^x/4^x) = 9^x/4^x

=> 1+(6^x/4^x) = 9^x/4^x

It can be arranged as

=> 9^x/4^x = (6^x/4^x)+1

we know that

(a/b)^m= a^m /b^m

=> (9/4)^x = (6/4)^x +1

=> (3²/2²)^x = [(3/2)^x] + 1

=> [(3/2)²]^x =[(3/2)^x] + 1

We know that

(a^m)^n = a^mn

=> (3/2)^(2x) = (3/2)^x +1

=> [(3/2)^x]^2 = (3/2)^x +1

Put (3/2)^x = a then

=> a² = a+1

=> a²-a = 1

=> a²-(2/2) a = 1

=> a²-2(a)(1/2) = 1

On adding (1/2)² both sides then

=>a²-2(a)(1/2)+(1/2)² = 1+(1/2)²

=> [a-(1/2)]² = 1+(1/4)

Since (a-b)² = a²-2ab +b²

=> [a-(1/2)]² = (4+1)/4

=> [a-(1/2)]² = 5/4

=> a-(1/2) =±√(5/4)

=> a-(1/2) =±√5/2

=> a =(1/2)±√5/2

=> a = (1±√5)/2

=> a = (1+√5)/2 or (1-√5)/2

=> (3/2)^x = (1+√5)/2 or (3/2)^x = (1-√5)/2

On taking logarithms both sides then

=> log (3/2)^x = log (1+√5)/2

=> x log (3/2) = log (1+√5)/2

since log a^m = m log a

=> x =[ log (1+√5)/2 ]/[log (3/2)] and

x = [ log (1-√5)/2 ]/[log (3/2)]

Answer:-

The value of x for the given problem is

[log (1±√5)/2 ]/[log (3/2)]

Used formulae:-

• (a/b)^m= a^m /b^m
• (a-b)² = a²-2ab +b²
• (a^m)^n = a^mn
• log a^m = m log a