A logarithm answers the question: to what power must we raise a base to get a given number? It is the inverse operation of exponentiation. On the YÖS exam (MAT-2) logarithm problems test the definition, algebraic properties, change of base, exponential and logarithmic equations, and the decimal logarithm (characteristic and mantissa). Almost every problem reduces to a short chain of standard transformations, so the key is to know the basic identities perfectly and always keep track of the domain: the argument of a logarithm must be positive, and the base must be positive and different from $1$.
For $a>0$, $a\ne 1$ and $b>0$ the logarithm of $b$ to base $a$ is the exponent $c$ such that $a^c=b$:$$\log_a b = c \iff a^c = b.$$Immediate consequences: $\log_a 1 = 0$, $\log_a a = 1$, $\log_a a^k = k$ and the fundamental identity $a^{\log_a b} = b$.Typical uses of the definition:1) $\log_4 2 = \frac{1}{2}$, because $4^{1/2} = \sqrt{4} = 2$.2) $\log_6(3x+1) = 0 \Rightarrow 3x+1 = 6^0 = 1 \Rightarrow x = 0$.3) $\log(x+22) = 2$ (base $10$) $\Rightarrow x+22 = 10^2 = 100 \Rightarrow x = 78$.Whenever an equation has the form $\log_a(\ldots) = c$, rewrite it at once as the argument being equal to $a^c$.
The expression $\log_a f(x)$ is defined only when all three conditions hold:$$f(x) > 0, \qquad a > 0, \qquad a \ne 1.$$If the base also depends on $x$, e.g. $\log_{g(x)} f(x)$, require $f(x)>0$, $g(x)>0$ and $g(x)\ne 1$ simultaneously.On YÖS this matters in two ways: (1) before solving a logarithmic equation, write down the domain; (2) after solving, substitute each root back and discard any value that makes an argument non-positive or a base equal to $1$. For example, in $\log_3(\log(8x)) - \log_3(\log(10x-1)) = 0$ we need $8x>0$ and $10x-1>0$, and additionally $\log(8x)>0$ and $\log(10x-1)>0$, i.e. $8x>1$ and $10x-1>1$, because these inner logarithms are themselves arguments of $\log_3$. Skipping this check is the most common source of wrong answers.
For $a>0$, $a\ne1$ and positive $x$, $y$:$$\log_a(xy) = \log_a x + \log_a y, \qquad \log_a\frac{x}{y} = \log_a x - \log_a y,$$$$\log_a x^k = k\log_a x, \qquad \log_{a^m} x = \frac{1}{m}\log_a x.$$Combining the last two rules gives the most useful YÖS formula: the exponent of the argument goes to the numerator, the exponent of the base goes to the denominator:$$\log_{a^m} b^n = \frac{n}{m}\log_a b.$$Examples: $\log_{25} 9 = \log_{5^2} 3^2 = \frac{2}{2}\log_5 3 = \log_5 3$; $\quad \log_5 \sqrt{5} = \log_5 5^{1/2} = \frac{1}{2}$; $\quad \log_{\sqrt7} 7 = \log_{7^{1/2}} 7 = 2$. Hence $\frac{\log_5\sqrt5}{\log_{\sqrt7}7} = \frac{1/2}{2} = \frac{1}{4}$.
Change of base formula (any $c>0$, $c\ne1$):$$\log_a b = \frac{\log_c b}{\log_c a}, \qquad \log_a b = \frac{1}{\log_b a}.$$It gives the chain (telescoping) rule, which collapses long products:$$\log_a b \cdot \log_b c = \log_a c.$$Example: $\log_2 5 \cdot \log_5 7 \cdot \log_7 8 = \log_2 8 = 3$, because each intermediate base cancels.If bases and arguments carry powers, first pull the exponents out with $\log_{a^m} b^n = \frac{n}{m}\log_a b$, then telescope. This is exactly the mechanism behind products such as $\log_{25}9\cdot\log_{27}16\cdot\log_4 5$ (see the worked example below). A convenient universal tactic: convert everything to natural logarithms via $\log_a b = \frac{\ln b}{\ln a}$ and cancel the fractions.
Two special bases have their own notation: the natural logarithm $\ln x = \log_e x$ (base $e \approx 2.718$) and the common (decimal) logarithm $\log x = \log_{10} x$. Useful values: $\ln e = 1$, $\log 10 = 1$, $\log 100 = 2$, $\log 0.01 = -2$.Any common logarithm can be split as$$\log x = n + m,$$where the integer $n$ is the characteristic and $m \in [0,1)$ is the mantissa. If $\log x = 3.162453$, the characteristic is $3$ and the mantissa is $0.162453$; the number $x$ has $3+1 = 4$ digits before the decimal point.Key trick with negatives: $\log\frac{1}{x} = -\log x$. For $\log x = 3.162453$ we get $\log\frac{1}{x} = -3.162453 = -4 + 0.837547$, so the characteristic is $-4$ and the mantissa is $0.837547$ (the mantissa must stay in $[0,1)$, so we borrow one unit from the integer part). Similarly, $-\log(x^2) = 2.44$ gives $2\log x = -2.44$, i.e. $\log x = -1.22 = -2 + 0.78$.
Three standard situations appear on YÖS.1) Both sides reduce to the same base: $a^{f(x)} = a^{g(x)} \Rightarrow f(x) = g(x)$.2) The equation is linear or quadratic in $a^x$: substitute $t = a^x$, $t>0$. Remember the shift rule $a^{x\pm k} = a^x \cdot a^{\pm k}$.Example: $18\cdot2^x + 5\cdot2^{x-1} - 164 = 0$. Since $2^{x-1} = \frac{2^x}{2}$, set $t = 2^x$: $18t + \frac{5t}{2} = 164 \Rightarrow \frac{41t}{2} = 164 \Rightarrow t = 8 = 2^3$, so $x = 3$.3) If bases cannot be matched, take a logarithm of both sides: $e^{x-1} = 3 \Rightarrow x - 1 = \ln 3 \Rightarrow x = 1 + \ln 3$. In general $a^{f(x)} = b \Rightarrow f(x) = \log_a b$. Always discard roots with $t \le 0$ after the substitution $t = a^x$, because an exponential is strictly positive.
Standard techniques:1) $\log_a f(x) = c \Rightarrow f(x) = a^c$ (definition).2) $\log_a f(x) = \log_a g(x) \Rightarrow f(x) = g(x)$, valid only inside the domain, so check $f(x)>0$ and $g(x)>0$ afterwards.3) If the equation is polynomial in $\log_a x$, substitute $t = \log_a x$.Nested example: $\log_3(\log(8x)) - \log_3(\log(10x-1)) = 0$. Move one term to the right: $\log_3(\log(8x)) = \log_3(\log(10x-1))$, so $\log(8x) = \log(10x-1)$, so $8x = 10x - 1$, giving $x = \frac{1}{2}$. Check the domain: $8x = 4 > 1$ so $\log(8x) = \log 4 > 0$, and $10x-1 = 4 > 1$ so $\log(10x-1) > 0$ — every logarithm is defined, the root is valid.Rule of thumb: solve first, then verify each candidate against all original logarithms; extraneous roots typically appear when arguments were multiplied or squared.
Symmetry trick. For positive $a$, $b$:$$a^{\ln b} = b^{\ln a},$$because $\ln(a^{\ln b}) = \ln b \cdot \ln a = \ln a \cdot \ln b = \ln(b^{\ln a})$. The same holds for any base: $a^{\log_c b} = b^{\log_c a}$.Application: $3^{\ln x} + 2x^{\ln 3} - 9 = 0$. Since $x^{\ln 3} = 3^{\ln x}$, set $t = 3^{\ln x} > 0$: $t + 2t - 9 = 0 \Rightarrow t = 3$. Then $3^{\ln x} = 3^1 \Rightarrow \ln x = 1 \Rightarrow x = e$.Taking logs. When the unknown sits both in the base and in the exponent, like $x^{\log_2 x} = 32x^4$, take $\log_2$ (or $\ln$) of both sides; the equation becomes quadratic in $\log_2 x$. After the substitution $t = \log_2 x$, Vieta's formulas give useful shortcuts: if $t_1 + t_2 = s$, then the product of the roots in $x$ equals $x_1 x_2 = 2^{t_1}\cdot 2^{t_2} = 2^{s}$.
Sums of logarithms collapse into the logarithm of a product, and vice versa. Two YÖS patterns:1) Product of powers. $\log_3(3^1\cdot 3^2\cdot \ldots \cdot 3^n) = \log_3 3^{1+2+\ldots+n} = \frac{n(n+1)}{2}$. If this equals $15$, then $\frac{n(n+1)}{2} = 15 \Rightarrow n(n+1) = 30 \Rightarrow n = 5$.2) Telescoping sum. Since $1 + \frac{1}{k} = \frac{k+1}{k}$:$$\sum_{k=1}^{n}\log_2\left(1+\frac{1}{k}\right) = \log_2\left(\frac{2}{1}\cdot\frac{3}{2}\cdot\ldots\cdot\frac{n+1}{n}\right) = \log_2(n+1),$$because everything cancels except the first denominator and the last numerator. For instance, the sum up to $k = 31$ equals $\log_2 32 = 5$.Strategy: whenever you see a sum of logarithms with the same base, merge it into one logarithm of a product and look for cancellation.
Compute $\log_{25}9\cdot\log_{27}16\cdot\log_4 5$.Step 1. Pull the exponents out of each factor using $\log_{a^m} b^n = \frac{n}{m}\log_a b$:$$\log_{25}9 = \log_{5^2}3^2 = \log_5 3, \qquad \log_{27}16 = \log_{3^3}2^4 = \frac{4}{3}\log_3 2, \qquad \log_4 5 = \log_{2^2}5 = \frac{1}{2}\log_2 5.$$Step 2. Collect the constants and telescope the chain:$$\frac{4}{3}\cdot\frac{1}{2}\cdot\log_5 3\cdot\log_3 2\cdot\log_2 5 = \frac{2}{3}\cdot\log_5 5 = \frac{2}{3}.$$Here $\log_5 3\cdot\log_3 2 = \log_5 2$ and then $\log_5 2\cdot\log_2 5 = 1$.Answer: $\frac{2}{3}$.
Solve $18\cdot2^x+5\cdot2^{x-1}-164=0$.Step 1. Express every power through $2^x$: $2^{x-1} = \frac{2^x}{2}$.Step 2. Substitute $t = 2^x$ (with $t > 0$):$$18t + \frac{5t}{2} - 164 = 0 \Rightarrow \frac{36t + 5t}{2} = 164 \Rightarrow 41t = 328 \Rightarrow t = 8.$$Step 3. Return to $x$: $2^x = 8 = 2^3$, hence $x = 3$.Step 4. Check: $18\cdot 8 + 5\cdot 4 - 164 = 144 + 20 - 164 = 0$. Correct.Answer: $x = 3$. The same pattern with base $e$: $e^{x-1} = 3 \Rightarrow x - 1 = \ln 3 \Rightarrow x = 1 + \ln 3$.
Find the product of the roots of $x^{\log_2 x}=32x^4$ (domain: $x>0$).Step 1. Take $\log_2$ of both sides:$$\log_2 x \cdot \log_2 x = \log_2 32 + 4\log_2 x \Rightarrow (\log_2 x)^2 = 5 + 4\log_2 x.$$Step 2. Substitute $t = \log_2 x$:$$t^2 - 4t - 5 = 0 \Rightarrow (t-5)(t+1) = 0 \Rightarrow t_1 = 5,\ t_2 = -1.$$Step 3. Return to $x$: $x_1 = 2^5 = 32$, $x_2 = 2^{-1} = \frac{1}{2}$; both are positive, so both are valid.Step 4. Product of the roots: $x_1 x_2 = 2^{5}\cdot2^{-1} = 2^{4} = 16$. Shortcut: by Vieta $t_1 + t_2 = 4$, so $x_1x_2 = 2^{t_1+t_2} = 2^4 = 16$.Answer: $16$.
Given $\log x = 3.162453$, find the characteristic and mantissa of $\log\frac{1}{x}$.Step 1. Use the power rule: $\log\frac{1}{x} = -\log x = -3.162453$.Step 2. The mantissa must lie in $[0,1)$, so we cannot read off $-3$ and $-0.162453$ directly. Borrow one unit from the integer part:$$-3.162453 = -4 + (1 - 0.162453) = -4 + 0.837547.$$Step 3. Therefore the characteristic is $-4$ and the mantissa is $0.837547$.The same borrowing idea works in reverse problems: if $-\log(x^2) = 2.44$, then $2\log x = -2.44$, so $\log x = -1.22 = -2 + 0.78$: characteristic $-2$, mantissa $0.78$, and $x = 10^{-1.22}$.
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