How to solve 4^-x = x (4^-x)^1/-x = x^1/-x => 4 = x^1/-x (1) let: 1/x = a, so -a = -1/x, x=1/a. (2) put (2) in (1) => 4 = (1/a)^-a, 4 = (1/a)^-1xa, 4 = [(1/a)^-1]^a, note: (1/a)^-1 = a so: 4 = [(1/a)^-1]^a = a^a 2^2=a^a => a=2 put a value 2 in (2): 1/x = a => 1/x = 2 => x = 1/2 x = 1/2 is the answer
1/4^x=x. , x4^x=1 taking ln
lnx4^x=0. , lnx=-2xln2
lnx/x=-2ln2=2ln2^-1=ln2^-1/2^-1
result. , x=2^-1=1/2
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4^(-1)= (1/2)^2=x^(1/x)
x=1/2
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0,5.
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@SALogics Thank You!
4 to power of-x =1/4 to power of x, x equals 1/2
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How to solve 4^-x = x
(4^-x)^1/-x = x^1/-x
=> 4 = x^1/-x (1)
let: 1/x = a, so -a = -1/x, x=1/a. (2)
put (2) in (1) => 4 = (1/a)^-a, 4 = (1/a)^-1xa, 4 = [(1/a)^-1]^a,
note: (1/a)^-1 = a
so: 4 = [(1/a)^-1]^a = a^a
2^2=a^a => a=2
put a value 2 in (2): 1/x = a => 1/x = 2 => x = 1/2
x = 1/2 is the answer
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4^-×=×...=>2^-2×=×
=>(2)ln2=(2×)ln2 .e^ln2(2×)
=>W{(1)ln2,e^ln2(1)}=W{2×ln2.e^ln2(2×)}
=>1)ln2=2×.ln2
=>2×=1&×=1/2
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A Nice Exponential Equation: 4^(- x) = x; x =?
[4^(- x)]^(- 1/x) = x^(- 1/x), x^(- 1/x) = 4 = 2^2 = 2^[(- 1)(- 2)] = [2^(- 1)]^(- 2)
x^(- 1/x) = 4 = [2^(- 1)]^(- 2) = (1/2)^[- 1/(1/2)]; x = 1/2
Answer check:
x = 1/2: 4^(- x) = 4^(- 1/2) = 2^(- 1) = 1/2 = x; Confirmed
Final answer:
x = 1/2
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X=1/2, Explain later
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x = 1/2
Yes, you are right! ❤