What is the answer to log_{10}(x)-log_{10}(2)=1+log_{10}(21-x) ?

The answer to log_{10}(x)-log_{10}(2)=1+log_{10}(21-x) is x=20

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log_{10}(x)-log_{10}(2)=1+log_{10}(21-x)

lo g_{10} (x) - lo g_{10} (2) = 1 + lo g_{10} (21 - x)

x = 20

Solution steps

lo g_{10} (x) - lo g_{10} (2) = 1 + lo g_{10} (21 - x)

Add

lo g_{10} (2)

to both sides

lo g_{10} (x) - lo g_{10} (2) + lo g_{10} (2) = 1 + lo g_{10} (21 - x) + lo g_{10} (2)

Simplify

lo g_{10} (x) = 1 + lo g_{10} (21 - x) + lo g_{10} (2)

Apply log rules

x = 20 (21 - x)

Solve

x = 20 (21 - x) : x = 20

x = 20

Verify Solutions:

x = 20

True

The solution is

x = 20

lo g_{3} (x) = - 1

lo g_{4} (x) + lo g_{10} (8 x) = 1

lo g_{10} (x + 1) = lo g_{10} (2 x + 1)

lo g_{10} (x + 2) - lo g_{10} (x - 1) = 4

x^{1 + l o g_{10} (x)} = 10 x

What is the answer to log_{10}(x)-log_{10}(2)=1+log_{10}(21-x) ?
The answer to log_{10}(x)-log_{10}(2)=1+log_{10}(21-x) is x=20
Solve for x: log_{10}(x)-log_{10}(2)=1+log_{10}(21-x)
The solution is x=20