Algebra 2 (only 10 problems, please help!)?

Question

1.Solve: log(sub4) (x+6)-log(sub4) x>2

2.Find the first 6 terms of the sequence in which a(sub1)=7 and a(sub n+1)=a(sub n)+2(sub n), n_>(greater than or equal to) 1.

3. Evaluate: log(sub3) (1/243)

4. Solve: 2e (to the 4x power) + 2=9

5. Find the first three iterates for the function f(x)= (x-2)/2 for an initial value of x(sub0)=1

6. Find the first three iterates of the function for the given initial value: f(x)=3x(2 power)-6, x(sub0)=1.

7. Find the sum of the first 18 terms of the arithmetic series in which a(sub1)=5 and d=6

8. Solve: 5 (2d-9 power) _> (greater than or equal to) 20.

9.Solve: e(-3x power)_>(greater than or equal to)14.

10. Evaluate the expression: ln e(5.7 power)

Answer 1

I'll do some. Then you do the rest.

1.) log(sub 4)(x+6) - log(sub 4) x > 2. This expression implies the following:

(x + 6 / x) > (4)²
(x + 6) > (4)²x
(x + 6) > 16 x

Now we simply solve for x:

(x + 6) > 16 x
6 > 15 x
6/15 > x
2/5 > x or x < 2/5

Try it! It works.

3.) 243 = 3^5. So 1/243 = 1/3^5 = 3^(-5). That implies this:

log (sub 3)(1/243) = log (sub 3)(3^(-5) = -5

4.) 2e^(4x) + 2 = 9
First subtract 2 from both sides:
2e^(4x) = 9 - 2
2e^(4x) = 7

Take the natural log of both sides:

ln 2 + ln e^(4x) = ln 7
ln 2 + 4x = ln 7

Isolate the variable term:

4x = ln 7 - ln 2
4x = 1.945910 - 0.693147
4x = 1.252673
x = 1.252673/4
x = 0.313168 This is close. Sub it back into the equation, and see what results.

7.) Last term, l = a + (n -1) d
l = 5 + ( 18 - 1) 6
l = 5 + (17 * 6)
l = 5 + 102
l = 107

Sum of terms, S = n/2 (a + l):

S = 18/2 (5 + 107)
S = 9 (112)
S = 1008

8.) 5^(2d-9) ≥ 20

Take the log, base 10, of both sides:

2d - 9 (log 5) ≥ log (4 * 5)
2d - 9 (log 5) ≥ log 4 + log 5

Subtract log 5 from both sides:

2d - 9 (log 5) - log 5 ≥ log 4 + 0
2d - 9 -1 (log 5) ≥ log 4
2d - 9 -1 (log 5) ≥ log 4
2d - 10 (log 5) ≥ log 4

log 5 = 0.698970
log 4 = 0.602060

Divide both sides by log 5:

2d - 10 ≥ log 4 / log 5
2d - 10 ≥ 0.602060 / 0.698970
2d - 10 ≥ 0.861353
2d ≥ 0.861353 + 10
2d ≥ 10.861353
d ≥ 10.861353 / 2
d ≥ 5.430678

Check it by substituting the value in for d. It works.

9.) e^(-3x) ≥ 14

Take the natural log of both sides:

ln e^(-3x) ≥ ln 14
-3x ≥ ln 14
-3x / -3 ≤ (1/-3)(ln 14) Do you see why I reversed the ≥ sign?
x ≤ (1/-3)(ln 14)
ln 14 ~ 2.639057
x ≤ (-1/3)(2.639057)
x ≤ -0.879686

10.) ln e^5.7 = 5.7 ln e = 5.7 (1) = 5.7