What is the answer to log_{10}(20+y)+log_{10}(y)=2 ?

The answer to log_{10}(20+y)+log_{10}(y)=2 is y=10(sqrt(2)-1)

Solve for y: log_{10}(20+y)+log_{10}(y)=2

The solution is y=10(sqrt(2)-1)

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lo g_{10} (20 + y) + lo g_{10} (y) = 2

y = 10 (2 - 1)

Decimal Notation

y = 4.14213 \dots

Solution steps

lo g_{10} (20 + y) + lo g_{10} (y) = 2

Apply log rules

(20 + y) y = 100

Solve

(20 + y) y = 100 : y = 10 (2 - 1), y = - 10 (1 + 2)

y = 10 (2 - 1), y = - 10 (1 + 2)

Verify Solutions:

y = 10 (2 - 1)

True

, y = - 10 (1 + 2)

False

The solution is

y = 10 (2 - 1)

x, 2 ln (8 x + 9) - 1 = - 5

lo g_{49} (x) = - \frac{1}{2}

y, x = lo g_{10} (y) + 5

ln (x^{2} - 24) = ln (10) + ln (x)

x = 1 0^{l o g_{10} (\frac{1}{x})}

What is the answer to log_{10}(20+y)+log_{10}(y)=2 ?
The answer to log_{10}(20+y)+log_{10}(y)=2 is y=10(sqrt(2)-1)
Solve for y: log_{10}(20+y)+log_{10}(y)=2
The solution is y=10(sqrt(2)-1)