Part of the Oxford MAT Livestream

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These are the solutions for the Logarithms and powers worksheet.

Here's a copy of what was written on-screen during the livestream

The identity log(ax)=log(x)+log(a) Notes on logarithms. The identity log_a(x)=log_b(x) / log_b(a) is stated and proved by writing a=b^d and a^c=x, then taking logs of everything. Solutions to first two questions are annotated onto the question sheet. See below for solutions. Solution for third question (see below) is annotated onto question sheet, and a sketch graph is drawn for the fourth question.


  • Simplify $\log_{10} 3+\log_{10} 4$ into a single term.
    • This is $\log_{10} 12$.
  • If we write $a=\ln 2$ and $b=\ln 5$, then write the following in terms of $a$ and $b$. \[\ln 1024, \quad \ln 40, \quad \ln \sqrt{2/5}, \quad \ln 1/10, \quad \ln 1.024.\]
    • $\ln 1024=\ln\left(2^{10}\right)=10\ln 2=10a$.
    • $\ln 40=\ln 8+\ln 5=3a+b$.
    • $\ln \sqrt{2/5}=\frac{1}{2}\ln 2/5=\frac{1}{2}\left(a-b\right)$.
    • $\ln 1/10=-\ln 10=-\ln2-\ln 5=-a-b$.
    • $\ln 1.024=\ln 1024+\ln 1/1000=10a +3(-a-b)=7a-3b$.
    • There are other solutions, partly because $b=a\times\log_2 5$.
  • Expand $\left(e^x+e^{-x}\right)\left(e^y-e^{-y}\right)+\left(e^x-e^{-x}\right)\left(e^y+e^{-y}\right)$.
    • $e^{x+y}+e^{y-x}-e^{x-y}-e^{-x-y}+e^{x+y}-e^{y-x}+e^{x-y}-e^{-x-y}$.
    • That's $2e^{x+y}-2e^{-x-y}$.
  • Expand $\left(e^x+e^{-x}\right)\left(e^y+e^{-y}\right)+\left(e^x-e^{-x}\right)\left(e^y-e^{-y}\right)$.
    • $e^{x+y}+e^{y-x}+e^{x-y}+e^{-x-y}+e^{x+y}-e^{y-x}-e^{x-y}+e^{-x-y}$.
    • That's $2e^{x+y}+2e^{-x-y}$.
  • Solve $2^x=3$. Solve $0.5^x=3$. Solve $4^x=3$.
    • $2^x=3$ is what it means for $x$ to be $\log_2 3$.
    • If $0.5^x=3$ then $2^{-x}=3$ so $x=-\log_2 3$. Alternatively, just write down $x=\log_{0.5}3$.
    • If $4^x=3$ then $2^{2x}=3$ so $x=\frac{1}{2}\log_2 3$. Alternatively, just write down $x=\log_4 3$.
  • For which values of $x$ (if any) does $1^x=1$? For which values of $x$ (if any) does $1^x=3$?
    • $1^x=1$ is true for all real $x$.
    • $1^x$ is never equal to 3 for real $x$.

MAT questions

MAT 2015 Q1H

  • That's a very strange logarithm. The only way I can think of to get rid of it is to write \[4-5x^2-6x^3=(x^2+2)^2.\]
  • Now this is a polynomial question (but we should be careful to go back and check our solutions in the original equation).
  • The polynomial rearranges to $x^4+6x^3+9x^2=0$.
  • Either $x=0$ or $x^2+6x+9=0$ which has a repeated root $x=-3$.
  • Let's quickly check these solutions $\log_2(4)=2$ and $\log_{11}(4-45+162)=\log_{11}121=2$.
  • There are two distinct solutions.
  • The answer is (c).

MAT 2017 Q1I

  • Let's simplify each term. $\log_b\left(\left(b^x\right)^x\right)=x\log_b \left(b^x\right)=x^2\log_b b =x^2$.
  • Also, $\log_a\left(\frac{c^x}{b^x}\right)=x\log_a(c/b)=x\left(\log_a c-\log_a b\right)$.
  • And the last term is $-\log_a b\log_a c$.
  • So this is a quadratic. Even better, we can factorise it \[x^2+\left(\log_a c-\log_a b\right)x-\log_a b\log_a c=\left(x-\log_a b\right)\left(x+\log_a c\right).\]
  • Those roots are the same number if $\log_a b=-\log a_c$, which happens when $c=1/b$.
  • The answer is (d).

MAT 2013 Q1F

  • The only way that we can have $\log_b a=2$ is if $a=b^2$. Similarly, we must have $c-3=b^3$ and $c+5=a^2$.
  • These simultaneous equations aren't linear, but we can do our best to solve them.
  • For example, we must have $c+5=b^4$ and $c-3=b^3$, so $b^4-b^3=8$. How many solutions does that have?
  • Let's sketch $b^3(b-1)$ for $b>0$ (since we're told that $b$ is positive).

A polynomial starts at the origin, dips down below the x-axis, then passes through (1,0) and then increases, increasing faster and faster.

  • This is negative in between $0$ and $1$ and then after that it's positive and increases (because $(b-1)$ and $b^3$ are both increasing functions). It can only have one solution for $b^3(b-1)=8$. In fact, that solution is $b=2$, but what we're interested in for this question is that that's a unique solution.
  • Now look back at the other equations. We've got $a=b^2=4$ and $c=3+b^3=11$. Let's check the original logarithms. \[ \log_2 4=2,\qquad \log_2 (11-3)=3,\qquad \log_4 (11+5)=2\]
  • These equations specify $a$ uniquely (it's 4).
  • The answer is (a).

MAT 2013 Q1J

  • This is some new notation. It appears in the question in the expression $[2^x]$, which mean the largest integer less than or equal to $2^x$.
  • When $x=0$, the term $[2^x]$ is 1. It stays at 1 until we get to a point with $2^x=2$, because then the largest integer less than or equal to $2^x$ will be 2 instead. That happens when $x=1$. Then $2^x$ keeps increasing, but the largest integer under it is still 2, until we get to a point with $2^x=3$.
  • Time for a logarithm; that happens when $x=\log_2 3$. There's a pattern here; the function takes each value between 1 and $2^{n-1}$ (it doesn't quite get to $2^n$ until right at the end of the interval when $x=n$).

A staircase-shaped function with steps of different sizes. The first part of the graph is a flat line between x=0 and x=1 at y=1, then there is a shorter line from x=1 to x=log_2 3 at y=2, and then several more shorter lines at each integer value of y.

  • When we integrate this function from 0 to $n$, we're calculating the area under the graph. That's made of a series of rectangles, and the sum of the areas of those rectangles is \[ 1+(\log_2 3-1)\times 2+ (\log_2 4-\log_2 3)\times 3 +(\log_2 5-\log_2 4)\times 4+\dots + (n-\log_2(2^n-1))\times \left(2^n-1\right) \]
  • There's a pattern here (add some $\log_2 k$ then subtract slightly more $\log_2 k$) and it simplifies to \[ -1-\log_2 3-\log_2 4-\dots -\log_2 (2^n-1)+n(2^n-1)\]
  • That's $-\log_2\left(\left(2^n-1\right)!\right)+n\left(2^n-1\right)$ which is not quite one of the options.
  • But $\log_2 2^n=n$ so we can re-write this as $-\log_2 \left(\left(2^n\right)!\right)+n2^n$.
  • The answer is (b).


  • I really like logarithms, but in lots of these questions getting rid of the logarithm is a good thing to do!
  • In that last question, we simplified things by using laws of logarithms and also by using the $k!$ notation for $k\times(k-1)\times\dots\times 2\times1$, which is a very compact way to write that product.