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