Math 125 Test 5 S/01 (9 pt each)

Math 123c Sample Test 3

1. a. Write 5^-2=3D 1/25 in log form = b. Write log_b (a+c) =3D 2 in the exponential form

Ans: -2 =3D log₅(1/25) = &nb= sp; = Ans: a+c =3D b^{2 =}

2. Solve the following equations with= out the calculator (by using a common base):

a.= 5^x* 2= 5^-2x=3D 125 Ans: = x=3D -1 = b. 9^2x(3) =3D 81 = Ans: x=3D 3/4 = &nb= sp; =

3. a. Simplify log_b (b^-10.33<= span lang=3DES style=3D'mso-ansi-language:ES'>) Ans: -10.33 = b. Simplify e ^{LN( x- y)} Ans: x - y = &nb= sp; =

Solve the following equat= ions (without a calculator):

4. a<= /span>. log x(8)=3D -3/4 Ans: x=3D 1/16 = b. log₂₇ (9) =3D x = Ans: x=3D 2/3

c. log _x(9)=3D -2 Ans: x=3D 1/3 = d. log ₁₆ (= 4x + 2)=3D ½ &nbs= p; Ans: x=3D 1/2 =

Solve using a calculator, give the exact and approximate answers.

5. a. 3*10^2x+¹=3D 5 = &nb= sp; b. 9 ^{0.021x =} =3D 23

Ans: x=3D (log(5/3) – 1)/2= »-0.389 = Ans: x=3D (ln(23) /(ln(9)*0.021)»67.9 = &nb= sp;

c. e-0.05t = =3D12.44 = = &nb= sp;d. 3 + 2* 10^x=3D 5

Ans: x=3D -ln(12.44)/0.05»= -50.4 = Ans: x=3D 0

e. 5 – 2*<= span class=3DGramE>LN(2x –3) =3D 1.2 = &nb= sp; f. 5*LN(4x) + 12 =3D 0

Ans: x=3D (e^1.9+3)/2»4.84 = &nb= sp; Ans: x=3D e^-12/5/4»0.0226

g. 4²^+
x =3D 5^x

Ans: x=3D 2ln(4)/ (ln5 – Ln4) = »12.4

Solve without the calcula= tor:

6. a. log(5) - log(x-1) =3D log<= sub> (2) = &nb= sp; b. log₇ (x+5) + log₇(x-1) =3D1

Ans: x =3D7/2 = &nb= sp; = &nb= sp; Ans: x =3D2, (-6 in no good)

c.= log ₆(x) + log <= sub>6(x + 5) =3D 2 Ans: x =3D4, -9 is no good. = &nb= sp; =

Put the expression into a= single log:

7. -2*log(x) -3*log(y) - log(z)+ 7*log(w) = Ans:log

Split the following log i= n terms of individual logs

8. lo= g(_{) Ans: 3/2
*log(x) – 2 – ¼ * log(z)}

9. The formula for compound interest = is A=3DP*e^rt. How lon= g will it take for an principal of $3,000 to be five-fold = of the original deposit, becoming $15,000, if the interest rate is 6.5 %? Ans: 24.= 8 yr

10. In= flation makes the value of $1 worth less in the future. The formula for the value o= f $1 in t years in the future with inflation rate r is given by the formula is A= =3DP*e^-rt.

What i= s the (present) value of $1 in 10 years if the inflation rate is 4%? Ans: 67 cent= s

How many years will it ta= ke for $1 to depreciate to 50 cents? Ans: 17.3 yr

11. Suppose the half-life of a radioac= tive material is h. Starting with A_o amount of a, let A be the amount that is left after t years, then A =3D A_o2^-t/h. How long would it take for a radioactive material with half–life of 350 years to be reduced to only 20% of the starting amount? Ans: 812 yr

11. Identify the center and radius of each circle. Dra= w it and label the highest, lowest, leftmost and right most points.

a. (x &= #8211; 3)² + (y + 5)² =3D 9 = Ans: C(3, -5), r =3D3 = b. x ² + y = ² – 4x + 12y =3D 9 Ans: C= (2, -6), r =3D7

12. Find the center line, the vertex of the parabola

a. y =3D x²– 4x –21 = &nb= sp; = b. y² – x + 2y= =3D 4

Ans: V(2, -25), y-int =3D -= 21 = &nb= sp; = Ans: V(-4, -2), x-int =3D -4

x-int =3D 7, -3 = &nb= sp; = &nb= sp; y-int: (-2±Ö5)/2= »1.24, -3.24

13. Grap= h the following ellipses. Find the cente= r, label the highest, lowest, leftmost and right most points on the ellipse.

     a. _{=
            &nb=
sp;            =
            b. x²<=
/span>
– 2x + 4y²+ 8y =3D 11}

14.<= span style=3D'mso-spacerun:yes'> Graph the following hyperbolas. Find the center, draw the center box, label the vertices.

     a. _{=
            &nb=
sp;            =
            b. 4y²+ 8y – x² +=

2x =3D 11}