CMI BS Hons Chapter Practice

Check each statement one by one. 1. If $\log_3(x-2)=2$, then $\qquad x-2=3^2=9$ so $\qquad x=11$ This statement is true. 2. If $\log_2(x-1)=\log_2(5-x)$, then both arguments must be positive: $\qquad x-1>0,\ 5-x>0 \implies 1<x<5$ Since the bases are the same, $\qquad x-1=5-x$ so $\qquad 2x=6$ $\qquad x=3$ This satisfies the domain, so the statement is true. 3. $\log_a(u+v)=\log_a u+\log_a v$ is false in general. The correct product rule is $\qquad \log_a(uv)=\log_a u+\log_a v$ 4. If $\log_5(x^2-1)=0$, then $\qquad x^2-1=5^0=1$ so $\qquad x^2=2$ $\qquad x=\pm \sqrt{2}$ not $\pm 1$ Hence this statement is false. Therefore the true statements are $\boxed{A,B}$.

Write $\qquad \dfrac{8}{25}=2^3\cdot 5^{-2}$ Using the given property, $\qquad g\left(\dfrac{8}{25}\right)=g(2^3)+g(5^{-2})=3g(2)-2g(5)$ Now substitute: $\qquad 3g(2)-2g(5)=3\cdot 3-2(-1)=9+2=11$ Hence the answer is $\boxed{11}$.

First write the domain: $\qquad x-1>0,\ x-4>0 \implies x>4$ Now combine the logarithms: $\qquad \log_3\big((x-1)(x-4)\big) > 1$ Since the base $3>1$, the logarithm is increasing. Hence $\qquad (x-1)(x-4) > 3$ Expand: $\qquad x^2 - 5x + 4 > 3$ $\qquad x^2 - 5x + 1 > 0$ The roots of $x^2 - 5x + 1 = 0$ are $\qquad x = \dfrac{5 \pm \sqrt{21}}{2}$ So $\qquad x < \dfrac{5-\sqrt{21}}{2} \quad \text{or} \quad x > \dfrac{5+\sqrt{21}}{2}$ Now intersect with the domain $x>4$. Since $\qquad \dfrac{5+\sqrt{21}}{2} \approx 4.79$, the valid solution set is $\qquad x > \dfrac{5+\sqrt{21}}{2}$ Hence the least integer satisfying the inequality is $\boxed{5}$.

First write the domain. For the logarithms to exist, we need $\qquad x-1>0 \quad \text{and} \quad x-3>0$ So $\qquad x>3$ Now combine the logarithms: $\qquad \log_2\big((x-1)(x-3)\big)=3$ Convert to exponential form: $\qquad (x-1)(x-3)=2^3=8$ Expand: $\qquad x^2-4x+3=8$ $\qquad x^2-4x-5=0$ Factor: $\qquad (x-5)(x+1)=0$ So the candidate solutions are $\qquad x=5 \quad \text{or} \quad x=-1$ From the domain $x>3$, only $x=5$ is valid. Therefore, the answer is $\boxed{5}$.

Check each statement. 1. $\log_a(xy)=\log_a x+\log_a y$ is true. 2. $\log_a(x+y)=\log_a x+\log_a y$ is false. There is no such law. 3. $\log_a\left(\dfrac{x}{y}\right)=\log_a x-\log_a y$ is true. 4. $\log_a(x^r)=r\log_a x$ is true for positive $x$. Hence the true statements are $\boxed{A,C,D}$.

Check each statement. 1. $\log_a b=\dfrac{1}{\log_b a}$ is true. 2. $\log_a a=0$ is false; actually $\log_a a=1$. 3. $a^{\log_a x}=x$ is true for $x>0$. 4. $\log_a 1=0$ is true. Hence the correct answer is $\boxed{A,C,D}$.

We check each statement one by one. 1. If $a>1$, then $\log_a x$ is an increasing function. Hence from $0<u<v$ we get $\qquad \log_a u < \log_a v$ So statement 1 is true. 2. If $0<a<1$, then $\log_a x$ is a decreasing function. Hence from $0<u<v$ we get $\qquad \log_a u > \log_a v$ So statement 2 is false. 3. Solve $\log_{1/2}(x+1) > -1$. First the domain: $\qquad x+1>0 \Rightarrow x>-1$ Since the base $\dfrac12$ lies between $0$ and $1$, the inequality reverses when we remove the logarithm: $\qquad x+1 < \left(\dfrac12\right)^{-1}=2$ $\qquad x<1$ Combining with the domain: $\qquad -1<x<1$ So statement 3 is true. 4. Solve $\log_3(x-2)\ge 0$. Domain: $\qquad x-2>0 \Rightarrow x>2$ Since base $3>1$, the direction stays the same: $\qquad x-2\ge 3^0=1$ $\qquad x\ge 3$ This already satisfies the domain. So statement 4 is true. Hence the true statements are $1,3,4$. Therefore the correct answer is $\boxed{A,C,D}$.

Check each statement carefully. 1. $\log_2(x-3)\ge 0$ First, domain: $\qquad x-3>0 \implies x>3$ Since base $2>1$, the logarithm is increasing, so $\qquad x-3 \ge 2^0 = 1$ Hence $\qquad x \ge 4$ So statement 1 is true. 2. If $0<a<1$, then $\log_a x$ is decreasing, not increasing. Therefore $\qquad \log_a u > \log_a v \implies u < v$ So statement 2 is false. 3. For $\log_5(2x-1)$ to be defined, the argument must be positive: $\qquad 2x-1>0 \implies x>\dfrac12$ So statement 3 is true. 4. $\log_{1/2}(x-1)\le -1$ First, domain: $\qquad x-1>0 \implies x>1$ Since the base $\dfrac12$ lies between $0$ and $1$, the logarithm is decreasing, so the inequality reverses: $\qquad x-1 \ge \left(\dfrac12\right)^{-1} = 2$ Thus $\qquad x \ge 3$ So statement 4 is true. Hence the correct statements are $1,3,4$. Therefore, the answer is $\boxed{A,C,D}$.