logg(1/32). 14. log (1/27). 15. log In problems 17-28, solve the given equation for x wit 17. log, (27)–3. 20. log4 (x /3)–5/2. 23) log3 (2x-3)-2 26. 10g 12(2x-1)-2 18, log, (2) 3. 21. log32 (3x) -3/5. 24, log2(1-5x) 4. 27 log2(3×2+2x 1 In problems 13-32, graph the given function by us the graph of the basic functions F(x) l09x, F(x) 29. f(x)- logd-3. 2 2
The Correct Answer and Explanation is:
Problem Solutions:
13. log(1/32)\log(1/32)
log(1/32)=log(32−1)=−log(32)\log(1/32) = \log(32^{-1}) = -\log(32)
Since 32=2532 = 2^5,
−log(32)=−log(25)=−5log(2)≈−5(0.3010)=−1.505-\log(32) = -\log(2^5) = -5\log(2) \approx -5(0.3010) = -1.505
14. log(1/27)\log(1/27)
log(1/27)=−log(27)=−log(33)=−3log(3)≈−3(0.4771)=−1.4313\log(1/27) = -\log(27) = -\log(3^3) = -3\log(3) \approx -3(0.4771) = -1.4313
15. The question appears incomplete.
17. logb(27)=−3\log_b(27) = -3
This means:
b−3=27⇒1b3=27⇒b3=127⇒b=13b^{-3} = 27 \Rightarrow \frac{1}{b^3} = 27 \Rightarrow b^3 = \frac{1}{27} \Rightarrow b = \frac{1}{3}
18. logb(2)=3\log_b(2) = 3
So,
b3=2⇒b=23≈1.26b^3 = 2 \Rightarrow b = \sqrt[3]{2} \approx 1.26
20. log4(x/3)=−5/2\log_4(x/3) = -5/2
x/3=4−5/2=145/2=132⇒x=3/32x/3 = 4^{-5/2} = \frac{1}{4^{5/2}} = \frac{1}{32} \Rightarrow x = 3/32
21. log32(3x)=−3/5\log_{32}(3x) = -3/5
Rewrite:
3x=32−3/5=(25)−3/5=2−3=1/8⇒x=1/243x = 32^{-3/5} = (2^5)^{-3/5} = 2^{-3} = 1/8 \Rightarrow x = 1/24
23. log3(2x−3)=2⇒2x−3=32=9⇒x=6\log_3(2x – 3) = 2 \Rightarrow 2x – 3 = 3^2 = 9 \Rightarrow x = 6
24. log2(1−5x)=4⇒1−5x=24=16⇒x=−3\log_2(1 – 5x) = 4 \Rightarrow 1 – 5x = 2^4 = 16 \Rightarrow x = -3
26. log12(2x−1)=−2⇒2x−1=12−2=1/144⇒x=145/288\log_{12}(2x – 1) = -2 \Rightarrow 2x – 1 = 12^{-2} = 1/144 \Rightarrow x = 145/288
27. log2(3×2+2x+1)\log_2(3x^2 + 2x + 1)
This is not an equation, so no solution for xx; it may be for graphing.
29. f(x)=log(x−3)f(x) = \log(x – 3)
This is a transformation of the parent function f(x)=logxf(x) = \log x. It’s shifted 3 units to the right.
Explanation:
Logarithms are the inverse of exponentiation. The expression logb(a)=c\log_b(a) = c means bc=ab^c = a. They are used to solve exponential equations and simplify multiplicative processes. Common bases include base 10 (common logarithms) and base e≈2.718e \approx 2.718 (natural logarithms).
For logarithmic evaluations like Problems 13 and 14, we apply properties such as:
- log(1/a)=−log(a)\log(1/a) = -\log(a)
- log(ab)=blog(a)\log(a^b) = b\log(a)
This helps simplify logs with fractional arguments. For example, log(1/32)=−log(32)=−5log(2)\log(1/32) = -\log(32) = -5\log(2).
In solving equations like 17 to 26, we use the definition of logarithms to rewrite the equation in exponential form. For example, logb(a)=c\log_b(a) = c means bc=ab^c = a. Solving for xx often involves algebraic manipulation after converting to exponential form.
Domain is important: The argument of a logarithmic function must be positive. For instance, in log2(1−5x)=4\log_2(1 – 5x) = 4, we must check that 1−5x>0⇒x<0.21 – 5x > 0 \Rightarrow x < 0.2 — this confirms the solution x=−3x = -3 is valid.
For graphing functions like 29, we use transformations. The basic graph y=logxy = \log x is shifted right by 3 units to become y=log(x−3)y = \log(x – 3). This shift changes the vertical asymptote from x=0x = 0 to x=3x = 3.
Understanding these properties is essential in algebra, calculus, and applications like data modeling and sound intensity in physics.
