Find the value of $k$ and the value of $p$.

The line $L_2$ is tangent to the graph of $g$ at $\text{A}$ and has equation $y=\left(\text{ln}p\right)x+q+1$.

The line $L_2$ passes through the point $\left(-2\text{, }-2\right)$.

The gradient of the normal to $g$ at $\text{A}$ is $\frac{1}{\text{ln}\left(\frac{1}{3}\right)}$.

Find the equation of $L_1$ in terms of $x$.

Write down the coordinates of $\text{B}$.

Solve .

Solve the equation ${\log }_2(x+3)+{\log }_2(x-3)=4$.

Use the regression equation to estimate the value of y when x = 3.57.

Solve the simultaneous equations

$\text{lo}{\text{g}}_{2}6x=1+2\text{lo}{\text{g}}_{2}y$

$1+\text{lo}{\text{g}}_{6}x=\text{lo}{\text{g}}_6\left(15y-25\right)$.

Show that $\text{lo}{\text{g}}_{r^2}x=\frac{1}{2}\text{lo}{\text{g}}_{r}x$ where $r,x\in {\mathbb{R}}^{+}$.

Given that ${\log }_{10}\left(\frac{1}{2\sqrt{2}}\left(p+2q\right)\right)=\frac{1}{2}\left({\log }_{10}p+{\log }_{10}q\right),p>0,q>0$, find $p$ in terms of $q$.

Find the value of a and of b.

Given that $f^{\prime }\left(a\right)=\frac{1}{\text{ln}p}$, find the equation of $L_1$ in terms of $x$, $p$ and $q$.

Find the solution of ${\log }_{2}x-{\log }_{2}5=2+{\log }_{2}3$.

The relationship between x and y can be modelled using the formula y = kxⁿ, where k ≠ 0 , n ≠ 0 , n ≠ 1.

By expressing ln y in terms of ln x, find the value of n and of k.