Start with \(y=e^{ax+b}\).

Take natural logarithms of both sides:

\(\ln y=ax+b\).

Compare this with \(Y=mX+c\).

So a suitable choice is

\(Y=\ln y,\ X=x,\ m=a,\ c=b\).

Start with \(y=10^{ax-b}\).

Take logarithms base \(10\) of both sides:

\(\log_{10} y=ax-b\).

Compare this with \(Y=mX+c\).

So a suitable choice is

\(Y=\log_{10} y,\ X=x,\ m=a,\ c=-b\).

Start with \(y=ax^{-b}\).

Take natural logarithms:

\(\ln y=\ln(a x^{-b})\).

Use log laws:

\(\ln y=\ln a+\ln(x^{-b})\)

\(\ln y=\ln a-b\ln x\).

Compare with \(Y=mX+c\).

So a suitable choice is

\(Y=\ln y,\ X=\ln x,\ m=-b,\ c=\ln a\).

Start with \(y=ab^x\).

Take natural logarithms:

\(\ln y=\ln(ab^x)\).

Use log laws:

\(\ln y=\ln a+\ln(b^x)\)

\(\ln y=\ln a+x\ln b\).

Compare with \(Y=mX+c\).

So a suitable choice is

\(Y=\ln y,\ X=x,\ m=\ln b,\ c=\ln a\).

Start with \(a=e^{x^2+by}\).

Take natural logarithms:

\(\ln a=x^2+by\).

Rearrange to make \(y\) the subject:

\(by=-x^2+\ln a\)

\(y=-\dfrac{1}{b}x^2+\dfrac{\ln a}{b}\).

Compare with \(Y=mX+c\).

So a suitable choice is

\(Y=y,\ X=x^2,\ m=-\dfrac{1}{b},\ c=\dfrac{\ln a}{b}\).

Start with \(x^a y^b=8\).

Take natural logarithms:

\(\ln(x^a y^b)=\ln 8\).

Use log laws:

\(a\ln x+b\ln y=\ln 8\).

Rearrange for \(\ln y\):

\(b\ln y=-a\ln x+\ln 8\)

\(\ln y=-\dfrac{a}{b}\ln x+\dfrac{\ln 8}{b}\).

Compare with \(Y=mX+c\).

So a suitable choice is

\(Y=\ln y,\ X=\ln x,\ m=-\dfrac{a}{b},\ c=\dfrac{\ln 8}{b}\).

Start with \(x a^y=b\).

Take natural logarithms:

\(\ln x+\ln(a^y)=\ln b\).

So \(\ln x+y\ln a=\ln b\).

Rearrange for \(y\):

\(y=-\dfrac{1}{\ln a}\ln x+\dfrac{\ln b}{\ln a}\).

Compare with \(Y=mX+c\).

So a suitable choice is

\(Y=y,\ X=\ln x,\ m=-\dfrac{1}{\ln a},\ c=\dfrac{\ln b}{\ln a}\).

Start with \(y=a e^{-bx}\).

Take natural logarithms:

\(\ln y=\ln(a e^{-bx})\).

Use log laws:

\(\ln y=\ln a+\ln(e^{-bx})\)

\(\ln y=\ln a-bx\).

Compare with \(Y=mX+c\).

So a suitable choice is

\(Y=\ln y,\ X=x,\ m=-b,\ c=\ln a\).

Let \(\log_{10} y=mx+c\).

The gradient is

\(m=\dfrac{11-5}{6-2}=\dfrac64=\dfrac32\).

So \(\log_{10} y=\dfrac32x+c\).

Substitute the point \((2,5)\):

\(5=\dfrac32\cdot2+c=3+c\).

So \(c=2\).

Hence

\(\log_{10} y=\dfrac32x+2\).

Start from

\(\log_{10} y=\dfrac32x+2\).

Write both sides as a power of \(10\):

\(y=10^{\frac32x+2}\).

Now separate the powers:

\(y=10^2\cdot10^{\frac32x}\).

Since \(10^2=100\),

\(y=100\times10^{\frac32x}\).

Let \(\ln y=m\ln x+c\).

The gradient is

\(m=\dfrac{13-4}{5-2}=\dfrac93=3\).

So \(\ln y=3\ln x+c\).

Substitute the point \((2,4)\):

\(4=3\cdot2+c\).

So \(4=6+c\), hence \(c=-2\).

Therefore

\(\ln y=3\ln x-2\).

Start from

\(\ln y=3\ln x-2\).

Write both sides as powers of \(e\):

\(y=e^{3\ln x-2}\).

Separate the powers:

\(y=e^{-2}\cdot e^{3\ln x}\).

Since \(e^{3\ln x}=x^3\),

\(y=e^{-2}x^3=\dfrac{x^3}{e^2}\).

Start with

\(5^{2y}=3^{2x+1}\).

Take natural logarithms of both sides:

\(\ln(5^{2y})=\ln(3^{2x+1})\).

Use log laws:

\(2y\ln 5=(2x+1)\ln 3\).

Expand the right-hand side:

\(2y\ln 5=2x\ln 3+\ln 3\).

Now divide by \(2\ln 5\):

\(y=\dfrac{\ln 3}{\ln 5}x+\dfrac{\ln 3}{2\ln 5}\).

This is in the form \(Y=mX+c\) with \(Y=y\) and \(X=x\), so the graph of \(y\) against \(x\) is a straight line.

From the linearised form

\(y=\dfrac{\ln 3}{\ln 5}x+\dfrac{\ln 3}{2\ln 5}\),

the gradient is the coefficient of \(x\), namely

\(\dfrac{\ln 3}{\ln 5}\).

The line cuts the \(y\)-axis when \(x=0\), so

\(y=\dfrac{\ln 3}{2\ln 5}\).

Hence the intercept point is

\(\left(0,\dfrac{\ln 3}{2\ln 5}\right)\).

Start with

\(y=ax^n\).

Take natural logarithms of both sides:

\(\ln y=\ln a+\ln(x^n)\)

\(\ln y=n\ln x+\ln a\).

So, on a graph of \(\ln y\) against \(\ln x\), the gradient is \(n\) and the intercept is \(\ln a\).

Find the gradient:

\(n=\dfrac{4.02-3.22}{0.31-1.83}=\dfrac{0.80}{-1.52}\approx-0.5263\).

So \(n\approx-0.53\) to 2 significant figures.

Now find the intercept using \((0.31,4.02)\):

\(4.02=(-0.5263)(0.31)+c\)

\(c\approx4.1832\).

So \(\ln a\approx4.1832\).

Hence

\(a=e^{4.1832}\approx65.57\).

So \(a\approx66\) to 2 significant figures.

Therefore the required values are \(a=66\) and \(n=-0.53\) to 2 significant figures.

Start with

\(y=k e^{n(x-2)}\).

Take natural logarithms of both sides:

\(\ln y=\ln k+\ln\left(e^{n(x-2)}\right)\)

\(\ln y=\ln k+n(x-2)\)

\(\ln y=nx+\ln k-2n\).

So, on a graph of \(\ln y\) against \(x\), the gradient is \(n\) and the intercept is \(\ln k-2n\).

Find the gradient:

\(n=\dfrac{4.33-1.84}{7-1}=\dfrac{2.49}{6}=0.415\).

So \(n\approx0.42\) to 2 significant figures.

Now find the intercept using \((1,1.84)\):

\(1.84=0.415(1)+c\)

\(c=1.425\).

But \(c=\ln k-2n\), so

\(1.425=\ln k-2(0.415)\)

\(1.425=\ln k-0.83\)

\(\ln k=2.255\).

Hence

\(k=e^{2.255}\approx9.54\).

So \(k\approx9.5\) to 2 significant figures.

Therefore the required values are \(k=9.5\) and \(n=0.42\) to 2 significant figures.

Start with \(V=Ar^T\).

Take natural logarithms:

\(\ln V=\ln A+T\ln r\).

So a plot of \(\ln V\) against \(T\) is approximately a straight line with:

gradient \(=\ln r\), intercept \(=\ln A\).

From the fitted line, the gradient is about \(0.06\), so

\(r\approx e^{0.06}\approx1.06\).

This means an average annual return of about \(6\%\).

The intercept is about \(8.1\), so

\(A\approx e^{8.1}\approx3300\).

So the best estimate is \(A\approx3300\), \(r\approx1.06\), annual return \(\approx6\%\).

Start with \(P=kr^t\).

Take logarithms base \(10\):

\(\log P=\log k+t\log r\).

So a plot of \(\log P\) against \(t\) is approximately a straight line with:

gradient \(=\log r\), intercept \(=\log k\).

From the fitted line, the gradient is about \(0.11\), so

\(r\approx10^{0.11}\approx1.29\).

The intercept is about \(0.95\), so

\(k\approx10^{0.95}\approx8.9\).

So the best estimate is \(k\approx8.9\) and \(r\approx1.29\).

Start with \(5^y=6^{2x-5}\).

Take logarithms of both sides:

\(y\log 5=(2x-5)\log 6\).

Rearrange for \(y\):

\(y=\dfrac{2\log 6}{\log 5}x-\dfrac{5\log 6}{\log 5}\).

This is in the form \(y=mx+c\).

So the gradient is \(\dfrac{2\log 6}{\log 5}\), and the intercept is \(-\dfrac{5\log 6}{\log 5}\).

Start with \(y=Ak^x\).

Take natural logarithms:

\(\ln y=\ln A+x\ln k\).

So on a plot of \(\ln y\) against \(x\):

gradient \(=\ln k\), intercept \(=\ln A\).

From the best-fit line, the gradient is about \(0.4\), so

\(k\approx e^{0.4}\approx1.49\).

The intercept is about \(0.37\), so

\(A\approx e^{0.37}\approx1.45\).

So the best estimate is \(A\approx1.45\), \(k\approx1.49\).

Take logarithms of both sides:

\((x+2)\log 3=(x-1)\log 11\).

Expand:

\(x\log 3+2\log 3=x\log 11-\log 11\).

Collect the \(x\)-terms:

\(x(\log 3-\log 11)=-\log 11-2\log 3\).

So

\(x=\dfrac{-\log 11-2\log 3}{\log 3-\log 11}\approx3.53666\).

Therefore \(x\approx3.54\) to 3 significant figures.

Take natural logarithms of both sides:

\(2x=(x-3)\ln 5\).

Expand:

\(2x=x\ln 5-3\ln 5\).

Collect the \(x\)-terms:

\(2x-x\ln 5=-3\ln 5\).

So

\(x=\dfrac{-3\ln 5}{2-\ln 5}\approx-12.362474\).

Therefore \(x\approx-12.362\) to 3 decimal places.

Take natural logarithms of both sides:

\(x-1=(x+3)\ln 5\).

Expand:

\(x-1=x\ln 5+3\ln 5\).

Collect the \(x\)-terms:

\(x(1-\ln 5)=1+3\ln 5\).

So

\(x=\dfrac{1+3\ln 5}{1-\ln 5}\approx-9.56342\).

Therefore \(x\approx-9.56\) to 3 significant figures.