What is the answer to 2.02=-log_{10}(x) ?

The answer to 2.02=-log_{10}(x) is x= 1/(10^{2.02)}

Solve for x: 2.02=-log_{10}(x)

The solution is x= 1/(10^{2.02)}

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2.02 = - lo g_{10} (x)

x = \frac{1}{1 0 ^{2.02}}

Decimal Notation

x = 0.00954 \dots

Solution steps

2.02 = - lo g_{10} (x)

Move

lo g_{10} (x)

to the left side

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

Move

2.02

to the right side

lo g_{10} (x) = - 2.02

Apply log rules

x = \frac{1}{1 0 ^{2.02}}

Verify Solutions:

x = \frac{1}{1 0 ^{2.02}}

True

The solution is

x = \frac{1}{1 0 ^{2.02}}

lo g_{\frac{2}{3}} (x) = 7

y, x = - lo g_{10} (y)

11 = ln (0^{2} + x)

lo g_{10} (x) + lo g_{10} (x + 99) = 2

((ln (y) - 1)^{2}) = 0

What is the answer to 2.02=-log_{10}(x) ?
The answer to 2.02=-log_{10}(x) is x= 1/(10^{2.02)}
Solve for x: 2.02=-log_{10}(x)
The solution is x= 1/(10^{2.02)}