What is the answer to log_{10}(16x)-log_{10}(2)=log_{10}(32) ?

The answer to log_{10}(16x)-log_{10}(2)=log_{10}(32) is x=4

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log_{10}(16x)-log_{10}(2)=log_{10}(32)

lo g_{10} (16 x) - lo g_{10} (2) = lo g_{10} (32)

x = 4

Solution steps

lo g_{10} (16 x) - lo g_{10} (2) = lo g_{10} (32)

Move

lo g_{10} (2)

to the right side

lo g_{10} (16 x) = lo g_{10} (32) + lo g_{10} (2)

Apply log rules

16 x = 64

Solve

16 x = 64 : x = 4

x = 4

Verify Solutions:

x = 4

True

The solution is

x = 4

lo g_{x} (\frac{1}{8}) = - 1.5

\frac{lo g _{10} ( 2 x )}{lo g _{10} ( 14 )} = 1

- lo g_{a} (x)

5 = x^{3} lo g_{1} (x)

y, x = - ln (y)

What is the answer to log_{10}(16x)-log_{10}(2)=log_{10}(32) ?
The answer to log_{10}(16x)-log_{10}(2)=log_{10}(32) is x=4
Solve for x: log_{10}(16x)-log_{10}(2)=log_{10}(32)
The solution is x=4