What is the answer to log_{10}(8x)-log_{10}(2)=log_{10}(32) ?

The answer to log_{10}(8x)-log_{10}(2)=log_{10}(32) is x=8

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log_{10}(8x)-log_{10}(2)=log_{10}(32)

lo g_{10} (8 x) - lo g_{10} (2) = lo g_{10} (32)

x = 8

Solution steps

lo g_{10} (8 x) - lo g_{10} (2) = lo g_{10} (32)

Move

lo g_{10} (2)

to the right side

lo g_{10} (8 x) = lo g_{10} (32) + lo g_{10} (2)

Apply log rules

8 x = 64

Solve

8 x = 64 : x = 8

x = 8

Verify Solutions:

x = 8

True

The solution is

x = 8

x, z = \frac{ln ( x - 7 )}{y}

lo g_{10} (x) + lo g_{10} (x + 48) = 2

3 lo g_{2} (x + 2) = 15

lo g_{x} (3) = 4

8 - ln^{3} (x) = 0

What is the answer to log_{10}(8x)-log_{10}(2)=log_{10}(32) ?
The answer to log_{10}(8x)-log_{10}(2)=log_{10}(32) is x=8
Solve for x: log_{10}(8x)-log_{10}(2)=log_{10}(32)
The solution is x=8