1 Solutions to Homework Exercises : Change of Base Handout

215204 1 Solutions to Homework Exercises : Change of Base Handout

Question 1

(a) We simply use the formula: log

211 =log311

log32 (b) log

57 =log27

log25 (c) This time, we use the laws of logarithms to do some simplifying after using the formula: log 1

38 =log8log13=log8log1-log3=log8-log3=-log8log3

(d) For this, we want to simplify before we use the formula. Werecognize that we can switch from base 6 to

base 31.1 log631= log316 =log56log531 (e) This is similar to the previous one. Notice that since log axis defined thenxmust be positive. And since log axis in the denominator, thenxcannot be 1, because logb1 = 0 for any baseb. Thereforex >0 and x?= 1, soxcan be the base of logarithms. We get: 1 logax= logxa=log2alog2x (f) Again, we simplify before changing to base 2. This time, we recognize the formlogba logbc= logca. And after we change to base 2, we use laws of logarithms to simplify. We get: log36 log16= log1636 =log236log216=log2(4×9)log224=(log24) + (log29)4 (log222) + (log232)

4=2 + 2log234=1 + log232=12+log232

1

2+?12?

log

23 =12+ log231/2=12+ log2⎷3

Question 2

(a) We use the change of base formula, changing to basee, i.e., to ln. log

542 =ln42

ln5

(b) This time, we switch the base before we change to natural logarithms. Of course, the base of the given

logarithm is 10.1 log23= log2310 =ln10ln23 2 (c) Now, we have a 5 multiplier to contend with: 5 log210= 5?1log210? = 5log2 = 5?ln2ln10? =5ln2ln10=ln25ln10=ln32ln10 (d) And this time the multiplier is disguised: log

3⎷

e= log3e1/2=?12? log

3e=?12??

lneln3? =?12?? 1ln3? =12ln3=1ln32=1ln9

Question 3

(a) This is easier if we switch from base 25 to base 5: log

255 =1

log525=1log552=12 (b) Again, it"s easiest to switch: log

642 =1

log264=1log282=12log28=12log223=12(3)=16

(c) We recognize that 81 is a power of 3 (i.e., is a power of 9, which is a power of 3) and so is 27, so we

change to base 3: log

8127 =log327

log381=log333log392=3log3(32)2=3log332(2)=3log334=34 (d) This time, the base which both 125 and 25 are powers of is 5: log

12525 =log525

log5125=log552log5(5)(25)=2log553=23 (e) Here, the base is 1

4= 2-2and the number whose logarithm we"re taking is 8 = 23. Therefore base 2 will

be useful. log

1/48 =log28

log214=log223log22-2=3-2=-32

Question 4

(a) We start by switching both to base 5, then simplify using laws of logarithms: 1 log25+1log35= log52 + log53 = log5(2×3) = log56 We can return this to a form similar to what we started with by switching to base 6: log56 =1 log65. (b) We use the same approach as in the previous question: 1 loge+1log2e= ln10 + ln2 = ln20 =1log20e (c) This time it will be more work, because we need to manipulate fractions: log

5x+ log25x=1

logx5+1logx25=1logx5+1logx52=1logx5+12logx5=22logx5+12logx5 3

2logx5=?32??

1logx5?

=?32? log

5x= log5x3/2= log5(x3)1/2= log5⎷x3

3

Question 5

Here, we switch to basexso that we can use the laws of logarithms to simplify, and thenuse the fact that

the numbers we"re dealing with are powers of 2. Notice that this time we know thatx >0 but not that

x?= 1. In the special case thatx= 1, we have log41+log321 = 0+0 = 0. For any other value ofx, we have:

log

4x+ log32x=1

logx4+1logx32=1logx22+1logx25=12logx2+15logx2=510logx2+210logx2 7

10logx2=?710??

1logx2?

=?710? log

2x= log2x7/10= log2x0.7

Of course, ifx= 1 thenx0.7= 1 as well and the answer will still be 0. So although the derivation of log

4x+ log32x= log2x0.7isn"t valid whenx= 1, we can overlook that because even for that case the

formula gives the right answer.

Question 6

(a)1log410+1log2510= log4 + log25 = log(4×25) = log100 = log102= 2 (b) 1 log236+1log336= log362 + log363 = log36(2×3) = log366 =1log636=1log662=12 (or simply realize that log

366 = log36⎷

36 = log36361/2=12)

Question 7

We start by switching to base 125, and then change to base 5, since that"s where we"re trying to get to:

1 log8125= log1258 =log58log5125=log523log553=3log523= log52 Another Approach:(Simplify before switching bases.) 1 log8125=1log853=13log85=?13??

1log85?

=?13? (log

58) =log5233=3log523= log52

1 Solutions to Homework Exercises : Change of Base Handout

Question 1

(a) We simply use the formula: log

211 =log311

log32 (b) log

57 =log27

log25 (c) This time, we use the laws of logarithms to do some simplifying after using the formula: log 1

38 =log8log13=log8log1-log3=log8-log3=-log8log3

(d) For this, we want to simplify before we use the formula. Werecognize that we can switch from base 6 to

base 31.1 log631= log316 =log56log531 (e) This is similar to the previous one. Notice that since log axis defined thenxmust be positive. And since log axis in the denominator, thenxcannot be 1, because logb1 = 0 for any baseb. Thereforex >0 and x?= 1, soxcan be the base of logarithms. We get: 1 logax= logxa=log2alog2x (f) Again, we simplify before changing to base 2. This time, we recognize the formlogba logbc= logca. And after we change to base 2, we use laws of logarithms to simplify. We get: log36 log16= log1636 =log236log216=log2(4×9)log224=(log24) + (log29)4 (log222) + (log232)

4=2 + 2log234=1 + log232=12+log232

1

2+?12?

log

23 =12+ log231/2=12+ log2⎷3

Question 2

(a) We use the change of base formula, changing to basee, i.e., to ln. log

542 =ln42

ln5

(b) This time, we switch the base before we change to natural logarithms. Of course, the base of the given

logarithm is 10.1 log23= log2310 =ln10ln23 2 (c) Now, we have a 5 multiplier to contend with: 5 log210= 5?1log210? = 5log2 = 5?ln2ln10? =5ln2ln10=ln25ln10=ln32ln10 (d) And this time the multiplier is disguised: log

3⎷

e= log3e1/2=?12? log

3e=?12??

lneln3? =?12?? 1ln3? =12ln3=1ln32=1ln9

Question 3

(a) This is easier if we switch from base 25 to base 5: log

255 =1

log525=1log552=12 (b) Again, it"s easiest to switch: log

642 =1

log264=1log282=12log28=12log223=12(3)=16

(c) We recognize that 81 is a power of 3 (i.e., is a power of 9, which is a power of 3) and so is 27, so we

change to base 3: log

8127 =log327

log381=log333log392=3log3(32)2=3log332(2)=3log334=34 (d) This time, the base which both 125 and 25 are powers of is 5: log

12525 =log525

log5125=log552log5(5)(25)=2log553=23 (e) Here, the base is 1

4= 2-2and the number whose logarithm we"re taking is 8 = 23. Therefore base 2 will

be useful. log

1/48 =log28

log214=log223log22-2=3-2=-32

Question 4

(a) We start by switching both to base 5, then simplify using laws of logarithms: 1 log25+1log35= log52 + log53 = log5(2×3) = log56 We can return this to a form similar to what we started with by switching to base 6: log56 =1 log65. (b) We use the same approach as in the previous question: 1 loge+1log2e= ln10 + ln2 = ln20 =1log20e (c) This time it will be more work, because we need to manipulate fractions: log

5x+ log25x=1

logx5+1logx25=1logx5+1logx52=1logx5+12logx5=22logx5+12logx5 3

2logx5=?32??

1logx5?

=?32? log

5x= log5x3/2= log5(x3)1/2= log5⎷x3

3

Question 5

Here, we switch to basexso that we can use the laws of logarithms to simplify, and thenuse the fact that

the numbers we"re dealing with are powers of 2. Notice that this time we know thatx >0 but not that

x?= 1. In the special case thatx= 1, we have log41+log321 = 0+0 = 0. For any other value ofx, we have:

log

4x+ log32x=1

logx4+1logx32=1logx22+1logx25=12logx2+15logx2=510logx2+210logx2 7

10logx2=?710??

1logx2?

=?710? log

2x= log2x7/10= log2x0.7

Of course, ifx= 1 thenx0.7= 1 as well and the answer will still be 0. So although the derivation of log

4x+ log32x= log2x0.7isn"t valid whenx= 1, we can overlook that because even for that case the

formula gives the right answer.

Question 6

(a)1log410+1log2510= log4 + log25 = log(4×25) = log100 = log102= 2 (b) 1 log236+1log336= log362 + log363 = log36(2×3) = log366 =1log636=1log662=12 (or simply realize that log

366 = log36⎷

36 = log36361/2=12)

Question 7

We start by switching to base 125, and then change to base 5, since that"s where we"re trying to get to:

1 log8125= log1258 =log58log5125=log523log553=3log523= log52 Another Approach:(Simplify before switching bases.) 1 log8125=1log853=13log85=?13??

1log85?

=?13? (log

58) =log5233=3log523= log52