What is ln(x)+ln(7)-ln(2)-2ln(4) ?

The solution to ln(x)+ln(7)-ln(2)-2ln(4) is ln((7x)/(32))

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simplify ln(x)+ln(7)-ln(2)-2ln(4)

simplify

ln (x) + ln (7) - ln (2) - 2 ln (4)

ln (\frac{7 x}{32})

Solution steps

ln (x) + ln (7) - ln (2) - 2 ln (4)

Apply log rule:

lo g_{c} (a) + lo g_{c} (b) = lo g_{c} (ab)

ln (x) + ln (7) = ln (7 x)

= ln (x \cdot 7) - ln (2) - 2 ln (4)

Apply log rule:

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

ln (7 x) - ln (2) = ln (\frac{7 x}{2})

= ln (\frac{x \cdot 7}{2}) - 2 ln (4)

Apply log rule:

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

2 ln (4) = ln (4^{2})

= ln (\frac{7 x}{2}) - ln (4^{2})

Apply log rule:

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

ln (\frac{7 x}{2}) - ln (4^{2}) = ln (\frac{\frac{7 x}{2}}{4 ^{2}})

= ln (\frac{\frac{x \cdot 7}{2}}{4 ^{2}})

Simplify

\frac{\frac{x \cdot 7}{2}}{4 ^{2}} : \frac{7 x}{32}

= ln (\frac{7 x}{32})

- lo g_{10} (5.33 \cdot 1 0^{- 6})

3.17 + lo g_{10} (\frac{3.4}{0.16})

lo g_{7} (\frac{f}{49})

ln ∣ x ∣ - ln ∣ x + 1 ∣

lo g_{10} (\frac{( x y ) ^{2}}{z ^{3}})

What is ln(x)+ln(7)-ln(2)-2ln(4) ?
The solution to ln(x)+ln(7)-ln(2)-2ln(4) is ln((7x)/(32))