What is the answer to ln^2(x-6)+ln^2(m+1)=3 ?

The answer to ln^2(x-6)+ln^2(m+1)=3 is

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ln^2(x-6)+ln^2(m+1)=3

ln^{2} (x - 6) + ln^{2} (m + 1) = 3

x = e^{3 - l n^{2} (m + 1)} + 6 {\frac{1}{e ^{3}} - 1 \leq m \leq e^{3} - 1}, x = e^{- 3 - l n^{2} (m + 1)} + 6 {\frac{1}{e ^{3}} - 1 \leq m \leq e^{3} - 1}

Solution steps

ln^{2} (x - 6) + ln^{2} (m + 1) = 3

Rewrite the equation with

ln (x - 6) = u

u^{2} + ln^{2} (m + 1) = 3

Solve

u^{2} + ln^{2} (m + 1) = 3 : u = 3 - ln^{2} (m + 1), u = - 3 - ln^{2} (m + 1)

u = 3 - ln^{2} (m + 1), u = - 3 - ln^{2} (m + 1)

Substitute back

u = ln (x - 6),

solve for

x

Solve

ln (x - 6) = 3 - ln^{2} (m + 1) : x = e^{3 - l n^{2} (m + 1)} + 6

Solve

ln (x - 6) = - 3 - ln^{2} (m + 1) : x = e^{- 3 - l n^{2} (m + 1)} + 6

x = e^{3 - l n^{2} (m + 1)} + 6, x = e^{- 3 - l n^{2} (m + 1)} + 6

Verify Solutions:

x = e^{3 - l n^{2} (m + 1)} + 6 {\frac{1}{e ^{3}} - 1 \leq m \leq e^{3} - 1}, x = e^{- 3 - l n^{2} (m + 1)} + 6 {\frac{1}{e ^{3}} - 1 \leq m \leq e^{3} - 1}

The solutions are

x = e^{3 - l n^{2} (m + 1)} + 6 {\frac{1}{e ^{3}} - 1 \leq m \leq e^{3} - 1}, x = e^{- 3 - l n^{2} (m + 1)} + 6 {\frac{1}{e ^{3}} - 1 \leq m \leq e^{3} - 1}

lo g_{10} (x \cdot 1 0^{- 3}) = 2.85

x (1 - ln (x)) = 0

lo g_{2} (x) = 3.6 \cdot 1 0^{13}

5 = \frac{1}{3} lo g_{10} (x) + 3

lo g_{x} (\frac{16}{625}) = 4

What is the answer to ln^2(x-6)+ln^2(m+1)=3 ?
The answer to ln^2(x-6)+ln^2(m+1)=3 is