solve 3^x-3^(x^(1/2))-72=0?

Question

i.e., solve 3 power x - 3 power sqrtx-72=0 algebraically without using numerical methods. any one can?

Answer 1

Given: (3^x) - (3^x^½) - 72 = 0

Let Z = x^½
Z² = X

(3^Z²) - (3^Z) - 72 = 0
(3^Z²) - (3^Z) - 2*6² = 0 Now tread the powers are a quadratic equation.
Z² - Z - 2 = 0

(Z - 2)(Z + 1) = 0
Z = 2 or Z = -1

But Z = X^½
→ 2 = X^½ or -1 = X^½
→ 4 = X or 1 = X

Try both values of X.

(3^1) - (3^1^½) - 72 = 3 - 1 -72 is not equal to zero.

(3^4) - (3^4^½) - 72 = 81 - 9 -72 = 0 Answer confirmed.
X = 4.

Answer 2

Put sq.rt.of 3^1/2x=y,then the equation will be y^2-y-72=0
=>(y-9)(y+8)=0=> y=-8 or 9 resubstituting 3^1/2x=3^2
1/2x=2 and therefore x=4 check 3^4-3^2-72=0

Answer 3

since 3^x-3^(x^(1/2))-72=0
then 3^x -81 - 3^ + 9 =0
so ( 3^x - 3^4 ) - (3^x^(1/2) - 3 ^2) =0
Since the difference between two quantities = 0 (1)
So each of them =0
So 3^x - 3^4 = 0 i.e x = 4
and 3^x^(1/2) = 3^2 i.e x^(1/2)= 2 then x = 4
Then the solution is x = 4 which verifies the condition (1)

Answer 4

let u=3^(x^(1/2)) then u^2 = 3^x
substitute so the equation can become
u^2 - u - 72 = 0
(u-9)(u+8) = 0
u-9 = 0 or u+8 = 0
u = 9 or u= -8
substitute back in for u
3^(x^(1/2)) = 9
log(3^(x^(1/2))) = log(9)
x^(1/2)log(3) = log(9)
x^(1/2) = log(9)/log(3)
x^(1/2) = 2
squaring both sides of the equation
x = 4

We can't get a solution for the -8 since we can't get a logarithm for a negative argument.

Answer 5

Mathtech is lucky... he finds the correct answer with the wrong math! There is no general algebraic solution for this type of problem.

For a similar equation, such as

2^x - 2^(x^(1/2)) - 272 = 0

mathtech's method would give x = 8, which is incorrect...

Answer 6

x=4 is an answer, but mathtech's approach is incorrect. If u=3^(sqrt(x)), we do not have u^2=3^x. The square doubles the exponent, it doesn't square the exponent. In this case, it is simply luck that sqrt(x)=x/2.

Answer 7

its an imaginary value .answer is 15.5+ i(3.96) OR 15.5 -i(3.96)