What is expand 10[log_{10}(I/(10^{-12)})] ?

The solution to expand 10[log_{10}(I/(10^{-12)})] is 10log_{10}(I)+120

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expand 10[log_{10}(I/(10^{-12)})]

expand

10 [lo g_{10} (\frac{I}{1 0 ^{- 12}})]

10 lo g_{10} (I) + 120

Solution steps

10 (lo g_{10} (\frac{I}{1 0 ^{- 12}}))

Remove parentheses:

(a) = a

= 10 lo g_{10} (\frac{I}{1 0 ^{- 12}})

\frac{I}{1 0 ^{- 12}} = 1 0^{12} I

= 10 lo g_{10} (1 0^{12} I)

Simplify

lo g_{10} (1 0^{12} I) : lo g_{10} (I) + 12

= 10 (lo g_{10} (I) + 12)

Apply the distributive law:

a (b + c) = ab + a c

a = 10, b = lo g_{10} (I), c = 12

= 10 lo g_{10} (I) + 10 \cdot 12

Multiply the numbers:

10 \cdot 12 = 120

= 10 lo g_{10} (I) + 120

(2 p - 0.8) (2 p - 0, 5)

(3 x - \frac{1}{3})^{3}

\frac{d ^{3}}{d x ^{3}} ((y) (5 x^{4} - x^{3} + 2 x))

\frac{2 x - 5}{( 2 x - 4 ) ( x - 4 ) ( x - 1 )}

12 (\frac{3 π ^{3}}{2 x ^{2}} - \frac{π ^{5}}{20 x ^{4}})

What is expand 10[log_{10}(I/(10^{-12)})] ?
The solution to expand 10[log_{10}(I/(10^{-12)})] is 10log_{10}(I)+120