What is the answer to log_{10}(x-2)+log_{10}(x-3)=log_{10}(2) ?

The answer to log_{10}(x-2)+log_{10}(x-3)=log_{10}(2) is x=4

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log_{10}(x-2)+log_{10}(x-3)=log_{10}(2)

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

x = 4

Solution steps

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

Apply log rules

(x - 2) (x - 3) = 2

Solve

(x - 2) (x - 3) = 2 : x = 4, x = 1

x = 4, x = 1

Verify Solutions:

x = 4

True

, x = 1

False

The solution is

x = 4

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

b, lo g_{3} (\frac{a}{b}) = lo g_{27} (a) + 1

lo g_{9} (x - 5) + lo g_{9} (x + 3) = 1

lo g_{a} (x) = \frac{1}{b \cdot lo g _{a} ( c )}

lo g_{2} (x + 5) = 3

What is the answer to log_{10}(x-2)+log_{10}(x-3)=log_{10}(2) ?
The answer to log_{10}(x-2)+log_{10}(x-3)=log_{10}(2) is x=4
Solve for x: log_{10}(x-2)+log_{10}(x-3)=log_{10}(2)
The solution is x=4