What is the answer to (log_{10}(10-x^2))/(log_{10)(5-2x)}=2 ?

The answer to (log_{10}(10-x^2))/(log_{10)(5-2x)}=2 is x=1

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(log_{10}(10-x^2))/(log_{10)(5-2x)}=2

\frac{lo g _{10} ( 10 - x ^{2} )}{lo g _{10} ( 5 - 2 x )} = 2

x = 1

Solution steps

\frac{lo g _{10} ( 10 - x ^{2} )}{lo g _{10} ( 5 - 2 x )} = 2

Multiply both sides by

lo g_{10} (5 - 2 x)

\frac{lo g _{10} ( 10 - x ^{2} )}{lo g _{10} ( 5 - 2 x )} lo g_{10} (5 - 2 x) = 2 lo g_{10} (5 - 2 x)

Simplify

lo g_{10} (10 - x^{2}) = 2 lo g_{10} (5 - 2 x)

Apply log rules

10 - x^{2} = (5 - 2 x)^{2}

Solve

10 - x^{2} = (5 - 2 x)^{2} : x = 3, x = 1

x = 3, x = 1

Verify Solutions:

x = 3

False

, x = 1

True

The solution is

x = 1

3 - lo g_{3} (p) = lo g_{p} (9)

800 = 24 x + ln (\frac{x}{45})

lo g_{2} (x) = 2022

y, x = 2 + ln (y + 3)

2 lo g_{10} (x) - lo g_{10} (3) + x = 0

What is the answer to (log_{10}(10-x^2))/(log_{10)(5-2x)}=2 ?
The answer to (log_{10}(10-x^2))/(log_{10)(5-2x)}=2 is x=1
Solve for x: (log_{10}(10-x^2))/(log_{10)(5-2x)}=2
The solution is x=1