What is ln(x+7)-2ln(x-3)+ln(7) ?

The solution to ln(x+7)-2ln(x-3)+ln(7) is ln((7(x+7))/((x-3)^2))

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simplify ln(x+7)-2ln(x-3)+ln(7)

simplify

ln (x + 7) - 2 ln (x - 3) + ln (7)

ln (\frac{7 ( x + 7 )}{( x - 3 ) ^{2}})

Solution steps

ln (x + 7) - 2 ln (x - 3) + ln (7)

Apply log rule:

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

2 ln (x - 3) = ln ((x - 3)^{2})

= ln (x + 7) - ln ((x - 3)^{2}) + ln (7)

Apply log rule:

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

ln (x + 7) - ln ((x - 3)^{2}) = ln (\frac{x + 7}{( x - 3 ) ^{2}})

= ln (\frac{x + 7}{( x - 3 ) ^{2}}) + ln (7)

Apply log rule:

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

ln (\frac{x + 7}{( x - 3 ) ^{2}}) + ln (7) = ln (7 \cdot \frac{x + 7}{( x - 3 ) ^{2}})

= ln (7 \cdot \frac{x + 7}{( x - 3 ) ^{2}})

Multiply fractions:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= ln (\frac{7 ( x + 7 )}{( x - 3 ) ^{2}})

x + ln (\frac{x ^{2}}{( x + 1 ) ^{6}})

f (m) = lo g_{\frac{4 π ( m ^{3} )}{3}} (2 m)

f (x) = ln (2 (ln (x) - 1))

20 lo g_{10} (\frac{0.05}{4 π \cdot 8 \cdot 1 0 ^{3}}) + 8 + 8

5 \cdot lo g_{10} (x) + 7 \cdot lo g_{10} (y)

What is ln(x+7)-2ln(x-3)+ln(7) ?
The solution to ln(x+7)-2ln(x-3)+ln(7) is ln((7(x+7))/((x-3)^2))