Exercise 1.5

1. Determine the value of the following:

(i) log255 (ii) log813 (iii) log2116

(iv) log71 (v) logxx (vi) log2512

(vii) log100.01 (viii) log23827 (ix) 22+log23

Instruction

(i) log255: Since 25 = , we have log255 = log525 = log55= × =

(ii) log813: Since 81 = 34, we have log813 = log343 = log33 = × =

(iii) log2116: Since 116 = 2−4 , we have log2116 = log22−4 =

(iv) log71: Any logarithm of 1 equals , so log71 =

(v) logxx: Since x = x12, we have logxx = logxx12 =

(vi) log2512: Since 512 = 29, we have log2512 = log229 =

(vii) log100.01: 0.01 = 10−2, we have x =

(viii) log282713: 82713 = So log282713 = log223 = log22 - log23 =

(ix) 22+log23: 22 × 2log23 = × =

2. Write the following expressions as log N and find their values:

(i) log 2 + log 5 (ii) log516 - log52 (iii) 3 log104

(iv) 2 log 3 - 3 log 2 (v) log 10 + 2 log 3 - log 2

Instruction

(i) log 2 + log 5 = = =

(ii) log216 - log22 = = = =

(iii) 3log104 = =

(iv) 2 log 3 - 3 log 2 = - = - =

(v) log 10 + 2 log 3 - log 2 = + -

= 1 + - log 2 = + = =

3. Evaluate each of the following in terms of x and y, if it is given that x = log23 and y = log25

(i) log215 (ii) log27.5 (iii) log260 (iv) log26750

Instruction

(i) log215: We have 15 = ×

log215 = = log23 + log25 = +

(ii) log27.5: We have 7.5 = =

log27.5 = = = + + = + +

(iii) log260: 60 = × × ×

log260 = log_2 (2 × 2 × 3 × 5) = + +

(iv) log26750: 6750 = × × = × ×

log26750 = + + = + +

4. Expand the following:

(i) log1000 (ii) log128625 (iii) logx2y3z4 (iv) logp2q3r4 (v) logx3y2

Instruction

(i) log1000: 1000 =

log1000 = = =

(ii) log128625: log128625 = + = + = +

(iii) logx3y2z4: logx3y2z4 = logx3 + logy2 + logz4 = + +

(iv) logp2q3r4: logp2q3r4 = + + = + +

(v) logx3y2: logx3y2 = = * = *

5. If x2 + y2 = 25xy, then prove that 2 log(x + y) = 3log3 + logx + logy.

Instruction

Given: x2 + y2 = 25xy. Adding 2xy to both sides, we get: =

x+y2 =

Taking log on both sides: logx+y2 = log33xy ⇒ = + +

2log(x + y) = + log x + log y

Hence proved.

6. If logx+y3 = 12 (log x + log y), then find the value of xy + yx.

Instruction

logx+y3 = = =

x+y3 = ⇒ x + y =

Squaring both sides: x+y2 =

x2 + 2xy + y2 = 9xy ⇒ x2 + y2 = ⇒ xy + yx =

7. If 2.3x = 0.23y = 1000, then find the value of 1x + 1y.

Instruction

From 2.3x = 1000: x·log(2.3) = log(1000) =

x =

From 0.23y = 1000: y =

Now, 0.23 = = ⇒ log(0.23) = + = +

Therefore: 1x - 1y = -

1x - 1y = log(2.3)/3 -

= log(2.3)/3 - log(2.3)/3 + =

8. If 2x+1 = 31−x then find the value of x.

Instruction

We have: 2x+1 = 31−x. Taking log of both sides: =

+ = +

⇒ xlog2 + xlog3 = log3 - log2 ⇒ (log2 + log3) =

9. Is (i) log 2 rational or irrational? Justify your answer. (ii) log 100 rational or irrational? Justify your answer.

Instruction

(i) log 2 rational or irrational?

If log 2 were rational, say log 2 = pq where p, q are integers, then: =

10pqq = 2q ⇒ = 2q

10p = 2q ⇒ =

Since the left side has only factor 5 and right side has only factor 2, the statement is not true unless both sides equal, which happens only when p = .

But then log 2 = , which is .

(ii) log 100 rational or irrational?

Justification: log 100 = log = log 10 = × = Since 2 is a number, log 100 is .

Chapter 1: Real Numbers

Glossary

Exercise 1.5

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Chapter 1: Real Numbers

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Glossary

Exercise 1.5