Appendix N: Derivation of the Logarithm Change of Base Formula
We set out to prove the logarithm change of base formula: logb x = loga x loga b. To do so we let y = logb x and apply these as exponents on the base.
MATHEMATICS 0110A CHANGE OF BASE Suppose that we have
Let y = logb a. Then we know that this means that by = a. We can take logarithms to base c
Change of Base
Logarithms - changing the base
This leaflet gives this formula and shows how to use it. A formula for change of base. Suppose we want to calculate a logarithm to base 2. The formula states.
mc logs
Lesson 5-2 - Using Properties and the Change of Base Formula
Common logarithin and natural logarithm functions are typically built into calculator systems. However it is possible to use a calculator to evaluate.
1 Solutions to Homework Exercises : Change of Base Handout
log 8 log 3. (d) For this we want to simplify before we use the formula. after we change to base 2
Sol ChangeBase
Change of Base Formula.pdf
The Change of Base Formula. Use a calculator to approximate each to the nearest thousandth. 1) log3. 3.3. 2) log2. 30. 3) log4. 5. 4) log2. 2.1. 5) log 3.55.
Change of Base Formula
Properties of Logarithms and Change of Base Theorem
Properties of Logarithms and Change of Base Theorem. Logarithmic Properties. 1. loga 1=0 EXAMPLE: Write the following expression as a single logarithm.
Props of Log and Change of Base
6.11 Notes – Change of base and log equations
Objectives: 1) Use common logs to solve equations. 2) Apply the change of base formula. 1).
day notes . notes change of base keyed
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Learning Targets: • Apply the properties of logarithms in any base. ⚫ Compare and expand logarithmic expressions. Use the Change of Base Formula. SUGGESTED
Change-of-Base Formula. For any logarithmic bases a and b and
Problem #1. Use your calculator to find the following logarithms. Show your work with Change-of-Base Formula. a) b). 2 log 10. 1. 3 log 9 c). 7 log 11.
Lecture
Question 1
(a) We simply use the formula: log211 =log311
log32 (b) log57 =log27
log25 (c) This time, we use the laws of logarithms to do some simplifying after using the formula: log 138 =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
12+?12?
log23 =12+ log231/2=12+ log2⎷3
Question 2
(a) We use the change of base formula, changing to basee, i.e., to ln. log542 =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: log3⎷
e= log3e1/2=?12? log3e=?12??
lneln3? =?12?? 1ln3? =12ln3=1ln32=1ln9Question 3
(a) This is easier if we switch from base 25 to base 5: log255 =1
log525=1log552=12 (b) Again, it"s easiest to switch: log642 =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: log8127 =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: log12525 =log525
log5125=log552log5(5)(25)=2log553=23 (e) Here, the base is 14= 2-2and the number whose logarithm we"re taking is 8 = 23. Therefore base 2 will
be useful. log1/48 =log28
log214=log223log22-2=3-2=-32Question 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: log5x+ log25x=1
logx5+1logx25=1logx5+1logx52=1logx5+12logx5=22logx5+12logx5 32logx5=?32??
1logx5?
=?32? log5x= log5x3/2= log5(x3)1/2= log5⎷x3
3Question 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 thatx?= 1. In the special case thatx= 1, we have log41+log321 = 0+0 = 0. For any other value ofx, we have:
log4x+ log32x=1
logx4+1logx32=1logx22+1logx25=12logx2+15logx2=510logx2+210logx2 710logx2=?710??
1logx2?
=?710? log2x= 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 log4x+ 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 log366 = 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? (log58) =log5233=3log523= log52
1 Solutions to Homework Exercises : Change of Base HandoutQuestion 1
(a) We simply use the formula: log211 =log311
log32 (b) log57 =log27
log25 (c) This time, we use the laws of logarithms to do some simplifying after using the formula: log 138 =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
12+?12?
log23 =12+ log231/2=12+ log2⎷3
Question 2
(a) We use the change of base formula, changing to basee, i.e., to ln. log542 =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: log3⎷
e= log3e1/2=?12? log3e=?12??
lneln3? =?12?? 1ln3? =12ln3=1ln32=1ln9Question 3
(a) This is easier if we switch from base 25 to base 5: log255 =1
log525=1log552=12 (b) Again, it"s easiest to switch: log642 =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: log8127 =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: log12525 =log525
log5125=log552log5(5)(25)=2log553=23 (e) Here, the base is 14= 2-2and the number whose logarithm we"re taking is 8 = 23. Therefore base 2 will
be useful. log1/48 =log28
log214=log223log22-2=3-2=-32Question 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: log5x+ log25x=1
logx5+1logx25=1logx5+1logx52=1logx5+12logx5=22logx5+12logx5 32logx5=?32??
1logx5?
=?32? log5x= log5x3/2= log5(x3)1/2= log5⎷x3
3Question 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 thatx?= 1. In the special case thatx= 1, we have log41+log321 = 0+0 = 0. For any other value ofx, we have:
log4x+ log32x=1
logx4+1logx32=1logx22+1logx25=12logx2+15logx2=510logx2+210logx2 710logx2=?710??
1logx2?
=?710? log2x= 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 log4x+ 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 log366 = 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? (log58) =log5233=3log523= log52
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