What is the answer to x^{1+log_{5}(x)}=25x^2 ?

The answer to x^{1+log_{5}(x)}=25x^2 is x=25,x= 1/5

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x^{1+log_{5}(x)}=25x^2

x^{1 + l o g_{5} (x)} = 25 x^{2}

x = 25, x = \frac{1}{5}

Decimal

x = 25, x = 0.2

Solution steps

x^{1 + l o g_{5} (x)} = 25 x^{2}

Apply log rules

(1 + lo g_{5} (x)) lo g_{5} (x) = lo g_{5} (25) + 2 lo g_{5} (x)

Rewrite the equation with

lo g_{5} (x) = u

(1 + u) u = lo g_{5} (25) + 2 u

Solve

(1 + u) u = lo g_{5} (25) + 2 u : u = 2, u = - 1

u = 2, u = - 1

Substitute back

u = lo g_{5} (x),

solve for

x

Solve

lo g_{5} (x) = 2 : x = 25

Solve

lo g_{5} (x) = - 1 : x = \frac{1}{5}

x = 25, x = \frac{1}{5}

Verify Solutions:

x = 25

True

, x = \frac{1}{5}

True

The solutions are

x = 25, x = \frac{1}{5}

lo g_{10} (x + 3) = 1 - lo g_{10} (x)

lo g_{x} (1) = \frac{1}{2}

lo g_{3} ((2 x - 3)^{2}) - 2 = 0

4.1 = - lo g_{10} (x)

ln (x + 1) - ln (x - 1) = 1

What is the answer to x^{1+log_{5}(x)}=25x^2 ?
The answer to x^{1+log_{5}(x)}=25x^2 is x=25,x= 1/5
Solve for x: x^{1+log_{5}(x)}=25x^2
The solution is x=25,x= 1/5