Rule: Ln(A/B) = Ln a Lnb.Answer Is Ln9-Ln 10 (FYI -.105)

rule:ln(a/b) = ln a – lnb.Answer is ln9-ln10 (FYI ≈-.105)

Answer is A

rule: log xxy=y so log226=6Answer is 6

rule: if log bx=y then by=x so b=5, x=M, y=5 and Answer is 55=M

rule: log (xy) = log x + log y so log (10,000x) = log 10,000 + log x. Answer is 4 + log x

rule: k ln x=lnxk and ln(a/b) = ln a – lnb so the first part becomes ln x4, the second becomes ln y1/6 and then combine. Answer is ln (x4/y1/6)

rule:log2(a/b) = log2 a – log2b so log21 -log28 = 0-3. Answer is -3

Graph them all or recognize that when x=0, y=0. By examining each equation, we see for x=0,
the y values are (1st column 1,-1,0,3) (2nd column 1,-1,1,0) so it’s the 3rd or 8th graph. For the 3rd, when x gets big, y gets big in the positive direction—not the example. For the 8th, when x gets big, y gets big in the negative direction. Graph #8 to be sure. Answer is G

rule: log (a/b) = log a - log b so log (y/100,000) = log y - log 100,000. Answer islog y - 5

rule: log 10x=x, so let 10log 48=y and take log of both sides. log 10log 48 = log y. The left side ismplybecomes:log 48, so log 48 = log y and therefore y=48 Answer is 48

rule: if log by=z then bz=y so b=4, y=x+2, z=3. So 43=x+2, 64=x+2, x=62Answer 62

rule: log (a/b) = log a - log b and lnyk= k lny so log4(16/√x+5) = log4 16 – log4 √x+5 =

log4 42 – log4 (x+5)1/2Answer is 2 - ½ log4 (x+5)

rule: log a - log b = log (a/b) so log510 – log52 = log5(10/2) = log55 = 1 Answer is 1

Graph them all or recognize that when x=0, y=1. By examining each equation, we see for x=0,
the y values are 1/3,1,0,3)so it’s the 2nd graph. Graph #2 to be sure.Answer is B

rule: k log x=logxk so 2log x=log x2 = log16 so x2 = 16 and x = ± 4 , but you can’t take the log of a negative number so x = -4 is not valid. Answer is A: 4

rule: log (xy) = log x + log y so log4(16x) = log416 + log4x = log442 + log4x. Answer is 2 + log4x

rule: lna + lnb = ln(ab) so ln(x-6) + ln(x+1) = ln((x-6)(x+1)) = ln(x2-5x-6) which also is ln(x-15) from the original equation. If ln(x2-5x-6) = ln(x-15) then (x2-5x-6) =(x-15), solve for x. x2-6x+9 = (x-3)2 so x = 3, but if x = 3 then two of the three original ln expressions:ln(x-6) and ln(x-15) are undefined, so there is no solution. Answer is B: There is no solution

rule: if log by=z then bz=y so b=3, y=x-4, z=-3. So 3-3 = x-4, 1/27=x-4, x=4+1/27. Answer is A: 109/27

note that 73=343: 72x-1 = 343 = 73 so 2x-1 = 3 and x = 2. Answer is 2

Graph them all or recognize thata transformation of the type g(x) = a + f(x) is simply f(x) raised by a at every point of x. A looks like a good candidate. Graph A to be sure. Answer for a) is A

The vertical asymptote for g(x) is the same as f(x) which is simply the y-axis or x=0. Answer for b) is 0

The domain of g(x) is the valid set of values on the x-axis where the function is defined. This is everything to the right of the y-axis or in interval notation (0,∞). Answer for c) is (0, ∞)

The range of g(x) is the valid set of values on the y-axis where the function is defined. The function is defined for all values or in interval notation (-∞, ∞). Answer for d) is (-∞, ∞)

rule: if log by=z then bz=y so b=2, y=3x+2, z=3. So 23= 3x+2, 8=3x+2, x=2.Answer is 2

note that 21/2 = √2: 2(x-4)/4 = √2 = 21/2 so (x-4)/4 = 1/2 and x-4 = 2 so x = 6. Answer is 6

rules: log (xy) = log x + log y, log (x/y)= log x - log y, logxk = k log x. note √a=a1/2 so rewriting:

log(a1/2 b5/c2) = log a1/2+log b5 - log c2 = ½ log a + 5 log b – 2 log c. Answer is ½ logfa + 5 logfb – 2 logfc

rules: logxk = k log x and a√x = x1/aln12√x = ln x1/12 = 1/12 lnxAnswer is 1/12 lnx

rule: logxk = k log xlogb x4 = 4logb xAnswer is 4logb x

Graph them all or recognize thata transformation of the type g(x) = f(x)-a is simply f(x) lowered by a at every point of x. C looks like a good candidate. Graph C to be sure. Answer is C

The domain of g(x) is the valid set of values on the x-axis where the function is defined. The function is definedfor all values or in interval notation (-∞, ∞). Answer is (-∞, ∞)

The range of g(x) is the valid set of values on the y-axis where the function is defined. This is everything above the horizontal asymptote at y=-3 or in interval notation (-3,∞). Answer is (-3, ∞)

The horizontal asymptote for g(x) is the same as f(x) which is simply the line y=-3. Answer is -3

The domain of logarithmic functions f(x)=ln(x) are all values of x that are larger than 0. In this case all the values of (x-1)2 that are larger than 0. So (x-1)2> 0 leads to (x-1) > 0 or x > 1.

In interval notation this is (1, ∞). Answer is (1, ∞)

rule: log x + log y = log (xy) so log3x + log3(2x-1) = log3(2x2-x) = 1 = log3 3 so (2x2-x) = 3 so 2x2-x-3 = 0

2x2-x-3 = (2x-3)(x+1) = 0 which means that x=3/2 or x=-1 but x must be > ½ to make (2x-1)>0 so the only value for x that is valid is x=3/2. Answer is A: 3/2

note that 2√4 = 41/2 and 4=22 so 2√4=2 leads to ½ log 4 = log 2.Answer is ½ log 4 = log 2

note: change of base means logbx = (logkx)/(logkb). Graph them all or recognize that when x=0, y=0. This means B or D could be valid except the (x+1) term means there will be a vertical asymptote at y=-1 so B is the answer. Graph B to be sure. Answer is B

4-2=1/16, 4-1=1/4, 40=1, 41=4, 42=16Answer is 1/16, ¼, 1, 4, 16

¾ -2= 16/9, ¾-1= 4/3, ¾ 0= 1, ¾ 1= ¾ , ¾ 2= 9/16Answer is 16/9, 4/3, 1, ¾, 9/16

The answer from Graphing calculator shows it is B

The domain of g(x) is the valid set of values on the x-axis where the function is defined. The function is defined for all values or in interval notation (-∞, ∞). Answer is (-∞, ∞)

The range of g(x) is the valid set of values on the y-axis where the function is defined. This is everything above the horizontal axis y=0 or in interval notation (0,∞). Answer is (0, ∞)

The horizontal asymptote for g(x) is the same as f(x) which is simply the x=axis or y=0. Answer is 0

Graph them all or recognize thata transformation of the type g(x) = a + f(x) is simply f(x) raised by a at every point of x. D looks like a good candidate. Graph D to be sure. Answer is D

The domain of g(x) is the valid set of values on the x-axis where the function is defined. The function is defined for all values or in interval notation (-∞, ∞). Answer is (-∞, ∞)

The range of g(x) is the valid set of values on the y-axis where the function is defined. This is everything above the line y=1 or in interval notation (1,∞). Answer is (1, ∞)

The horizontal asymptote for g(x) is the line y=1. Answer is 1

Graph them all or recognize thata transformation of the type g(x) = a + f(x) is simply f(x) raised by a at every point of x. B looks like a good candidate. Graph B to be sure. Answer is B

The domain of g(x) is the valid set of values on the x-axis where the function is defined. The function is defined for all values or in interval notation (-∞, ∞). Answer is (-∞, ∞)

The range of g(x) is the valid set of values on the y-axis where the function is defined. This is everything above the line y=1 or in interval notation (1,∞). Answer is (1, ∞)

The horizontal asymptote for g(x) is the line y=1. Answer is 1

5-2=1/25, 5-1=1/5, 50=1, 51=5, 52=25Answer is 1/25, 1/5, 1, 5, 25

1/3-2= 9, 1/3-1= 3, 1/30= 1, 1/31= 1/3 , 1/32= 1/9Answer is 9, 3, 1, 1/3 (0.33), 1/9 (0.11)

The answer is graph B and the intersection is at (0,4)

The Answer is C

The Answer is D