What is the answer to 2.12=-log_{10}(x) ?

The answer to 2.12=-log_{10}(x) is x= 1/(10^{2.12)}

Solve for x: 2.12=-log_{10}(x)

The solution is x= 1/(10^{2.12)}

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2.12 = - lo g_{10} (x)

x = \frac{1}{1 0 ^{2.12}}

Solution steps

2.12 = - lo g_{10} (x)

Move

lo g_{10} (x)

to the left side

2.12 + lo g_{10} (x) = 0

Move

2.12

to the right side

lo g_{10} (x) = - 2.12

Apply log rules

x = \frac{1}{1 0 ^{2.12}}

Verify Solutions:

x = \frac{1}{1 0 ^{2.12}}

True

The solution is

x = \frac{1}{1 0 ^{2.12}}

9000 = \frac{4000}{ln ( x + 10 )}

ln^{4} (x^{2}) = 16 ln (x)

lo g_{a} (x^{2}) = b

7 + lo g_{10} (x) = 10

5 ln (5 x - 9) + 11 = 26

What is the answer to 2.12=-log_{10}(x) ?
The answer to 2.12=-log_{10}(x) is x= 1/(10^{2.12)}
Solve for x: 2.12=-log_{10}(x)
The solution is x= 1/(10^{2.12)}