What is 3+10log_{10}(5)-40log_{10}(2) ?

The solution to 3+10log_{10}(5)-40log_{10}(2) is 3+log_{10}((9765625)/(2^{40)})

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simplify 3+10log_{10}(5)-40log_{10}(2)

simplify

3 + 10 lo g_{10} (5) - 40 lo g_{10} (2)

3 + lo g_{10} (\frac{9765625}{2 ^{40}})

Decimal Notation

- 2.05149 \dots

Solution steps

3 + 10 lo g_{10} (5) - 40 lo g_{10} (2)

Apply log rule:

a lo g_{c} (b) = lo g_{c} (b^{a})

10 lo g_{10} (5) = lo g_{10} (5^{10})

= 3 + lo g_{10} (5^{10}) - 40 lo g_{10} (2)

Apply log rule:

a lo g_{c} (b) = lo g_{c} (b^{a})

40 lo g_{10} (2) = lo g_{10} (2^{40})

= 3 + lo g_{10} (5^{10}) - lo g_{10} (2^{40})

Apply log rule:

lo g_{c} (a) - lo g_{c} (b) = lo g_{c} (\frac{a}{b})

lo g_{10} (5^{10}) - lo g_{10} (2^{40}) = lo g_{10} (\frac{5 ^{10}}{2 ^{40}})

= 3 + lo g_{10} (\frac{5 ^{10}}{2 ^{40}})

5^{10} = 9765625

= 3 + lo g_{10} (\frac{9765625}{2 ^{40}})

- lo g_{10} (4.6 \cdot 1 0^{- 10})

lo g_{e} (6) - lo g_{e} (2)

\frac{300}{2 \cdot π \cdot 1000} \cdot ln (\frac{4 + 2 \cdot 100 0 ^{2}}{25})

- \frac{1}{5} ln ∣ x + 1 ∣ + \frac{1}{5} ln ∣ x - 4 ∣ + C

ln (\frac{1}{x}) - ln (2)

What is 3+10log_{10}(5)-40log_{10}(2) ?
The solution to 3+10log_{10}(5)-40log_{10}(2) is 3+log_{10}((9765625)/(2^{40)})