4^x + 6^x=9^x
Help.

The value of x is {log (1±√5)/2 }/{log (3/2)}.

Step-by-step explanation:

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

On applying exponent radical rule of both sides then

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

Where, a = 9, b = 4 and m = x.

Similarly, apply this rule also on RHS.

Where, a = 6, b = 4 and m = x then

⇛(9/4)^x = (6/4)^x +1

⇛(3²/2²)^x = {(3/2)^x} + 1

⇛{(3/2)²}^x = {(3/2)^x} + 1

Again applying exponent rule on LHS.

Since, (a^m)^n = a^mn

Where, a = (3/2), m = 2 and n = x.

⇛(3/2)^(2x) = (3/2)^x +1

⇛{(3/2)^x}^2 = (3/2)^x +1

Put (3/2)^x = a then

⇛a² = a+1

Shift variable (a) RHS from LHS, changing it's sign.

⇛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)

⇛{a-(1/2)}² = (1/1) + (1/4)

Take the LCM of 1 and 4 is 4 on RHS.

⇛{a-(1/2)}² = {(1*4 + 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

Shift the fraction number -(1/2) from LHS to RHS, changing it's sign.

⇛a =(1/2)±√5/2

Take the LCM of 2 and 2 is 2 as common.

⇛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

On applying log a^m = m log a, we get

⇛x ={log (1+√5)/2 }/{log (3/2)} and

Therefore, x = {log (1-√5)/2 }/{log (3/2)}

Answer: Hence, the value of x for the given equation is {log (1±√5)/2 }/{log (3/2)}.

Please let me know if you have any other questions.

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