Laws of Logarithms
Write down the value of log5(1/125)
5? = 1/125
? = -3
Solve log8x = 1/3
81/2 = x
3√8 = x
2 = x
Find the sum of the infinite geometric series:
1/3 + 2/15 + 4/75 + ...
u1 = 1/3
r = [2/15]/[1/3] = 2/5
S∞ = [1/3]/[1-2/5]
S∞ = 5/9
Evaluate 11C9.
11C9 = 11!/[9!(11-9)!]
= [11*10*9!]/[9!*2!]
= (11*10)/2
= 11*5
= 55
Show that 1/3 - 1/5 = 2/15
Put both fraction on LHS (left hand side) in terms of fractions with 15 in the denominator:
5/15 - 3/15 = 2/15
5-3/15 = 2/15
2/15 = 2/15 ✓
Let a = log5b, where b>0. Write the following expression in terms of a.
log5b4
log5b4 = log5(5a)4 = log5(54a)
= 4a
50 = x + 3
1 = x + 3
-2 = x
Find the range of values of x for which the series converges:
1 + 2x + 4x2
-1 < r < 1
r = 2x/1 = 2x
2x = -1 2x = 1
x = -1/2 x = 1/2
-1/2 < x < 1/2
Consider the expansion of (2x-1)9.
How many terms are in the expansion?
10
Show that (2n−1)3+(2n+1)3=16n3+12n for n∈ℤ
[8n3-12n2+6n-1]+[8n3+12n2+6n+1] (expand both binomials)
adding them together, the -12n2 and +12n2 cancel as well as the -1 and +1 leaving
16n3+12n, for n∈ℤ
Let a = log5b, where b>0. Write the following expression in terms of a.
log525b
log5(52*5a) = log5(52)+log5(5a)
=2+a
Solve log10(x+5) + log102 = 3
log10(2(x+5)) = 3
103 = 2(x+5)
1000 = 2(x+5)
500 = x+5
495 = x
For the geometric series 2 + 4x + 8x2, find the expression, in terms of x, for the sum to infinity of the series.
S∞ = u1/1-r
r = u2/u1 = 4x/2 = 2x
S∞ = 2/1-2x
Consider the expansion of (2x-1)9.
Find the coefficient of the term in x2.
**CALCULATOR
9C7 *(2x)9-7 * (-1)7
= 36*4x*-1
= -144x2
the coefficient is -144
Find the value of a such that x3 - y3 ≡ (x - y)(x2+axy+y2)
multiply out right side to get:
x3 + ax2y + xy2 - x2y - axy2 - y3
in order for this to = (x3 - y3)
ax2y = -x2y and xy2 = -axy2
in both cases, a must be -1
a = -1
Let a = log5b, where b>0. Write the following expression in terms of a.
log25b
log25b = log5b/log525
= log5(5a)/2
= a/2
Solve log28+log5(1/25)+log93=log16x
Evaluate each log on the right
3 + (-2) + (1/2) = log16x
3/2 = log16x
163/2 = x
(42)3/2 = 43 = 64
x = 64
The second term of a geometric series is 2 and its sum to infinity is 9. Find two possible values of the common ratio.
**CALCULATOR
S∞ = u1/1-r
r = u2/u1 which means u1 = u2/r
9 = [2/r]/(1-r)
9 = 2/r(1-r)
9 = 2/(r-r2)
9(r-r2) = 2 [you could stop here and graph to find solutions]
-9r2+9r = 2
-9r2 + 9r - 2 = 0 (use quadratic formula)
r = 1/3, r = 2/3
Given that (x+2)n = xn + 18xn-1 + bxn-2, find the value of b.
**CALCULATOR
18 = nC1 * 2
9 = nC1, n = 9
9C2 * (2)2 = b
144 = b
Prove that the sum of the cubes of any two consecutive odd integers is divisible by four.
(2n−1)3+(2n+1)3=16n3+12n for n∈ℤ
factor 4n out of both terms on the right giving
16n3+12n = 4n(4n2+3), which is a multiple of 4
Show that log9(cos2x+2) = log3√cos2x+2
log9(cos2x+2) = log3(cos2x+2)/log39
= log3(cos2x+2)/2
= 1/2*log3(cos2x+2)
= log3(cos2x+2)1/2
= log3√cos2x+2
Solve the equation 52x+3 = 9x-5, giving your answer in terms of natural logarithms.
ln(52x+3) = ln(9x-5)
(2x+3)ln5 = (x-5)ln9
2xln5 + 3ln5 = xln9 - 5ln9
2xln5 - xln9 = -5ln9 - 3ln5
x(2ln5 - ln9) = -5ln9 - 3ln5
x = [-5ln9 - 3ln5]/[2ln5 - ln9] (you could factor -1 from top and bottom to clean it up)
An infinite geometric series has a sum to infinity of 27 and the sum of the first three terms is equal to 19. Find the first term.
**CALCULATOR
u1 + r*u1 + r2*u1 = 19
u1(1 + r + r2) = 19
u1 = 19/(1+r+r2)
27 = u1/1-r
27 = [19/(1+r+r2)]/(1-r)
27 = 19/[(1+r+r2)(1-r)
27 = 19/(1+r+r2-r-r2-r3)
27 = 19/(1-r3)
27(1-r3) = 19
27-27r3 = 19
-27r3 = -8
r3 = 8/27
r = 2/3
u1 = 19/(1+2/3+(2/3)2) (I used calc at this point)
u1 = 9
27 = 19/[1
The third term, in descending powers of x, in the expansion of (x+p)8 is 252x6. Find the possible values of p.
**CALCULATOR
252x6 = 8C2 * x8-2 * p2
252x6 = 28*x6*p2
252 = 28p2 (solve for p with calculator)
p = 3, p = -3
Prove algebraically that the difference between the squares of any two consecutive integers is always an odd number.
n2 + (n+1)2 = n2 + n2 + 2n + 1
= 2n2 + 2n + 1
= 2(n2 + n) + 1
= 2k + 1 where k is an integer
hence the sum is always odd