C3 2017 2 Edexcel

[CleanSlate]

[_gcx]

(a)

We have:

3x - 9 = \ln 8 = 3 \ln 2

So:

x = \dfrac 1 3 \left(3 \ln 2 + 9\right) = \ln 2 + 3

(b)

We have:

\ln(2y + 5) - \ln(4 - y) = 2

So:

\displaystyle \ln \left(\frac {2y + 5} {4 - y}\right) = 2

So:

\dfrac {2y + 5} {4 - y} = e^2

Then:

2y + 5 = 4e^2 - e^2 y

So:

(2 + e^2)y = 4e^2 - 5

giving:

\displaystyle y = \frac {4e^2 - 5} {e^2 + 2}

[CleanSlate]