What is ln(5e)-ln(10^3)+log_{2}(4^2) ?

The solution to ln(5e)-ln(10^3)+log_{2}(4^2) is ln(e/(200))+4

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simplify ln(5e)-ln(10^3)+log_{2}(4^2)

simplify

ln (5 e) - ln (1 0^{3}) + lo g_{2} (4^{2})

ln (\frac{e}{200}) + 4

Decimal Notation

- 0.29831 \dots

Solution steps

ln (5 e) - ln (1 0^{3}) + lo g_{2} (4^{2})

Apply log rule:

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

ln (5 e) - ln (1 0^{3}) = ln (\frac{5 e}{1 0 ^{3}})

= ln (\frac{5 e}{1 0 ^{3}}) + lo g_{2} (4^{2})

\frac{5 e}{1 0 ^{3}} = \frac{e}{200}

= ln (\frac{e}{200}) + lo g_{2} (4^{2})

Apply log rule

lo g_{a} (x^{b}) = b \cdot lo g_{a} (x),

assuming

x \geq 0

= ln (\frac{e}{200}) + 2 lo g_{2} (4)

2 lo g_{2} (4) = 4

= ln (\frac{e}{200}) + 4

3 ln (x - 8) - 4 ln (x)

ln_{e^{2}} (4 e^{8})

- lo g_{10} (1.24 \cdot 1 0^{- 3})

ln (18) - ln (30)

ln (\frac{1}{0.7})

What is ln(5e)-ln(10^3)+log_{2}(4^2) ?
The solution to ln(5e)-ln(10^3)+log_{2}(4^2) is ln(e/(200))+4