What is the answer to D=10log_{10}(1+a/r) ?

The answer to D=10log_{10}(1+a/r) is r=-a/(1-10^{\frac{D){10}}}

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D=10log_{10}(1+a/r)

D = 10 lo g_{10} (1 + \frac{a}{r})

r = - \frac{a}{1 - 1 0 ^{\frac{D}{10}}}

Solution steps

D = 10 lo g_{10} (1 + \frac{a}{r})

Switch sides

10 lo g_{10} (1 + \frac{a}{r}) = D

Divide both sides by

10

lo g_{10} (1 + \frac{a}{r}) = \frac{D}{10}

Apply log rules

1 + \frac{a}{r} = 1 0^{\frac{D}{10}}

Multiply both sides by

r

1 \cdot r + \frac{a}{r} r = 1 0^{\frac{D}{10}} r

Simplify

r + a = 1 0^{\frac{D}{10}} r

Solve

r + a = 1 0^{\frac{D}{10}} r : r = - \frac{a}{1 - 1 0 ^{\frac{D}{10}}}

r = - \frac{a}{1 - 1 0 ^{\frac{D}{10}}}

lo g_{10} (5 x + 20) = 3

x, y = ln (x)

3 lo g_{e} (x + 1) = - lo g_{e} (2 x - 1)

ln (x - 1) + ln (3) = 2

4 = \frac{ln ( 4 x )}{ln ( 2 x )}

What is the answer to D=10log_{10}(1+a/r) ?
The answer to D=10log_{10}(1+a/r) is r=-a/(1-10^{\frac{D){10}}}