Logarithm Identities, Change of Base & Sums

Definition. $\log_a N = x \iff a^x = N$, valid for base $a>0,\ a\neq 1$ and $N>0$. The three laws (same base throughout):

• Product: $\log_a(MN) = \log_a M + \log_a N$
• Quotient: $\log_a\!\dfrac{M}{N} = \log_a M - \log_a N$
• Power: $\log_a(M^k) = k\,\log_a M$

Special values: $\log_a a = 1$, $\log_a 1 = 0$, $\log_a(a^k) = k$, and $a^{\log_a N} = N$. Domain: you may only take the log of a positive number — the argument of every log in a problem must stay $>0$. This is what later forces solution-rejection.

Strategy for an evaluate-and-combine problem:

• Rewrite arguments as powers of the smallest convenient base: $27=3^3,\ 1024=2^{10},\ 3125=5^5$, and a nested radical like $\sqrt{7\sqrt{7\sqrt{7}}}$ as $7^{1/2+1/4+1/8}=7^{7/8}$.
• Pull exponents out front with the power law, then combine with product/quotient.
• Sum of logs across a list — e.g. $\sum_j \log_{10}(2^j5^j)=\sum_j j\,\log_{10}10=\sum_j j$ — telescopes once you notice $2\cdot 5 = 10$. Then finish the series: $\sum_{j=1}^{n}j=\dfrac{n(n+1)}{2}$, so "the sum equals 66" means $n(n+1)=132\Rightarrow n=11$, and "equals 15" means $n=5$. Also $\sum j^2=\dfrac{n(n+1)(2n+1)}{6}$ if the exponents are squares.

Change of base: $\log_b a = \dfrac{\log_c a}{\log_c b}$ for any valid base $c$. Two consequences carry most of the marks:

• Reciprocal identity: $\dfrac{1}{\log_a b} = \log_b a$ (set $c=a$). So $\dfrac{1}{\log_k N} = \log_N k$.
• Telescoping sum: $\displaystyle\sum_{k=2}^{m}\dfrac{1}{\log_k N} = \sum_{k=2}^{m}\log_N k = \log_N(2\cdot 3\cdots m) = \log_N(m!)$. When $N=m!$ this equals $\log_N N = 1$.
• Product collapse: $\log_a b\cdot\log_b a = 1$, and $\log_a b\cdot\log_b c = \log_a c$ (chain).

For base $a>1$:

• $\log_a N > 0 \iff N > 1$; $\quad\log_a N = 0 \iff N = 1$; $\quad\log_a N < 0 \iff 0 < N < 1$.
• The function $\log_a$ is strictly increasing, so $\log_a f(x)$ attains its minimum exactly where $f(x)$ is minimised (provided $f>0$ there).
• Strict monotonicity also proves UNIQUENESS: a strictly increasing left-hand side meets a constant at most once, so a root found by inspection is THE root. $5^x+4^x+3^x+2^x+1=5$ has $x=0$ by inspection, and because every term $a^x$ ($a>1$) is strictly increasing, the sum is too — there is exactly one solution. Without this line "$x=0$" is a guess; with it, it is a proof.

Minimising $\log_{10}(\text{quadratic}):$ complete the square — $x^2+bx+c = (x+\tfrac{b}{2})^2 + (c-\tfrac{b^2}{4})$ — the minimum argument is $c-\tfrac{b^2}{4}$, and the minimum of the log is $\log_{10}$ of that.

Two recurring shapes:

• AP / GP test on logs: $\ln x, \ln x^3, \ln x^5 = \ln x,\,3\ln x,\,5\ln x$. They are in AP (common difference $2\ln x$); for GP you must separately check $q^2 = pr$, i.e. $(3\ln x)^2 = (\ln x)(5\ln x)$ — here $9 \neq 5$, so never GP.
• Geometric mean of a list of powers: the GM of $1,2,2^2,\dots,2^{n-1}$ is $\big(2^{0+1+\cdots+(n-1)}\big)^{1/n} = 2^{(n-1)/2}$; taking $\log_2$ gives $\dfrac{n-1}{2}$, so expressions like $1+2\log_2 G$ simplify to $n$.
• An infinite GP hiding in the logs: $\log_e3+\log_{e^2}3+\log_{e^4}3+\dots=\ln3\,(1+\tfrac12+\tfrac14+\dots)$ by $\log_{a^k}N=\tfrac1k\log_aN$, and the bracket is $\dfrac{a}{1-r}=\dfrac{1}{1-\frac12}=2$ — the infinite-GP sum $S_\infty=\dfrac{a}{1-r}$ for $|r|<1$ (Sequence & Series). So the total is $2\ln3=\ln9$.