What is the answer to -log_{10}(b+98)=0.125 ?

The answer to -log_{10}(b+98)=0.125 is b= 1/(10^{0.125)}-98

Solve for b: -log_{10}(b+98)=0.125

The solution is b= 1/(10^{0.125)}-98

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- lo g_{10} (b + 98) = 0.125

b = \frac{1}{1 0 ^{0.125}} - 98

Decimal Notation

b = - 97.25010 \dots

Solution steps

- lo g_{10} (b + 98) = 0.125

Divide both sides by

- 1

lo g_{10} (b + 98) = - 0.125

Apply log rules

b + 98 = \frac{1}{1 0 ^{0.125}}

Solve

b + 98 = \frac{1}{1 0 ^{0.125}} : b = \frac{1}{1 0 ^{0.125}} - 98

b = \frac{1}{1 0 ^{0.125}} - 98

Verify Solutions:

b = \frac{1}{1 0 ^{0.125}} - 98

True

The solution is

b = \frac{1}{1 0 ^{0.125}} - 98

x, 2 y^{2} + 3 y - 5 x = ln (x + y^{2}) - 1

lo g_{7} (x) = lo g_{7} (a) \cdot b^{2} - lo g_{7} (a^{3} \cdot b)

- 4 = 20 \cdot lo g_{10} (\frac{x}{80})

ln (x^{2} + 6) - ln (x - 2) = ln (x + 1)

lo g_{10} (n) = 1.0043

What is the answer to -log_{10}(b+98)=0.125 ?
The answer to -log_{10}(b+98)=0.125 is b= 1/(10^{0.125)}-98
Solve for b: -log_{10}(b+98)=0.125
The solution is b= 1/(10^{0.125)}-98