What is the answer to log_{x}(5)-log_{25}(x)-1=0 ?

The answer to log_{x}(5)-log_{25}(x)-1=0 is x= 1/(5^{1+sqrt(3))},x=5^{sqrt(3)-1}

Solve for x: log_{x}(5)-log_{25}(x)-1=0

The solution is x= 1/(5^{1+sqrt(3))},x=5^{sqrt(3)-1}

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lo g_{x} (5) - lo g_{25} (x) - 1 = 0

x = \frac{1}{5 ^{1 + 3}}, x = 5^{3 - 1}

Decimal Notation

x = 0.01231 \dots, x = 3.24849 \dots

Solution steps

lo g_{x} (5) - lo g_{25} (x) - 1 = 0

Apply log rules

\frac{1}{2 lo g _{25} ( x )} - lo g_{25} (x) - 1 = 0

Rewrite the equation with

lo g_{25} (x) = u

\frac{1}{2 u} - u - 1 = 0

Solve

\frac{1}{2 u} - u - 1 = 0 : u = - \frac{1 + 3}{2}, u = \frac{3 - 1}{2}

u = - \frac{1 + 3}{2}, u = \frac{3 - 1}{2}

Substitute back

u = lo g_{25} (x),

solve for

x

Solve

lo g_{25} (x) = - \frac{1 + 3}{2} : x = \frac{1}{5 ^{1 + 3}}

Solve

lo g_{25} (x) = \frac{3 - 1}{2} : x = 5^{3 - 1}

x = \frac{1}{5 ^{1 + 3}}, x = 5^{3 - 1}

Verify Solutions:

x = \frac{1}{5 ^{1 + 3}}

True

, x = 5^{3 - 1}

True

The solutions are

x = \frac{1}{5 ^{1 + 3}}, x = 5^{3 - 1}

lo g_{10} (x) + lo g_{10} (x - 2) = 2 lo g_{10} (x - 4)

lo g_{10} (a) 5 = 5

- ln (5 - x) = 2.2

lo g_{3} (x) + lo g_{3} (x - 8) = 2

lo g_{2} (x + 1) = 3

What is the answer to log_{x}(5)-log_{25}(x)-1=0 ?
The answer to log_{x}(5)-log_{25}(x)-1=0 is x= 1/(5^{1+sqrt(3))},x=5^{sqrt(3)-1}
Solve for x: log_{x}(5)-log_{25}(x)-1=0
The solution is x= 1/(5^{1+sqrt(3))},x=5^{sqrt(3)-1}