Basic rules of logarithms
Basic rules of logarithms
You are already aware of the different rules for manipulating and simplifying expressions with indices or exponents, like the rule that says that x3 × x5 equals x8 due to the fact that you can add the exponents. In logarithm, there are similar rules.
Log Rules:
1) logb(mn) = logb(m) + logb(n)
2) logb(m/n) = logb(m) – logb(n)
3) logb(mn) = n x logb(m)
In informal terms, the log rules can be expressed as:
1) Multiplication within the log can be turned into addition outside the log, and vice versa.
2) Division within the log can be turned into subtraction outside the log, and vice versa.
3) An exponent on everything within a log can be moved out front as a multiplier, and vice versa.
The basic idea
A logarithm is the opposite of a an exponent or power. In other words, if we take a logarithm of a number, we undo an exponentiation. A logarithm is a function that does all this work for you. Just like we can alter the base b for the exponential function, we can as well alter the base b for the logarithmic function.
Basic rules for logarithms
Because taking a logarithm is the opposite of exponential function (more precisely, the logarithmic function log_b x is the inverse function of the exponential function b^x), we can derive the basic rules for logarithms from the basic rules of indices or exponents.
For ease of use, we'll write the rules in terms of the natural logarithm ln(x). The rules apply for any logarithm log_b x, apart from the fact that you ought to substitute any occurrence of e with the new base b.
The natural log was defined by equations. If we plug the value of k from equation \eqref{naturalloga} into equation \eqref{naturallogb}, we determine that a relationship between the natural log and the exponential function is e^{ln c} = c. We'll make use of equations {lnexpinversesa} and {lnexpinversesb} to derive the following rules for the logarithm.
Rule or special case Formula
Product ln(xy) = ln(x)+ln(y)
Quotient ln(x/y) = ln(x)-ln(y)
Log of power ln(x^y) = yln(x)
Log of e ln(e)=1
Log of one ln(1)=0
Log reciprocal ln(1/x)=-ln(x)
The product rule
We can use the product rule for exponentiation to derive a corresponding product rule for logarithms.
The quotient rule
The quotient rule for logarithms follows from the quotient rule for exponentiation, {e^a}{e^b} = e^{a-b} in a similar way.
Log of a power
To obtain the rule for the log of a power, we begin with the rule for power of a power, (e^a)^b = e^{ab}.
Beginning with c=x^y in equation and applying it again, this time only once more with c=x, we can calculate that e^{ln (x^y)} = x^y = e^{ln(x)})^y = e^{yln(x)} where in the last step we made use of the power of a power rule for a=ln(x) and b=y. From e^{ln (x^y)} = e^{yln(x)}, we can conclude that ln (x^y) = yln(x), which is the rule for the log of a power.
Log of e
The formula for the log of e comes from the formula for the power of one, e^1=e. .. Merely take the logarithm of both sides of this equation and use equation to conclude that ln(e) = 1.
Log of one
The formula for the log of one comes from the formula for the power of zero, e^0=1. merely take the logarithm of both sides of this equation and make use of the equation to conclude that ln(1) = 0.
Log of reciprocal
The rule for the log of a reciprocal follows from the rule for the power of negative one x^{-1} ={1}/{x} and the above rule for the log of a power. Merely substitute y=-1 into the log of power rule, and you have that ln(1/x) = -ln (x).
Expanding logarithms
Log rules can be used to simplify expressions, to "expand" expressions, or to solve for values.
Expand log3(2x).
When they say "expand", it means that you've been given one log expression with lots of stuff inside it, and you are required to make use of the logarithm rules to take the log apart into lots of different logs, each with only one thing inside. That is, you've been given one log with a composite argument, and they want you to convert this to a lot of logs, each with a simple argument.
I have a "2x" inside the log. Due to the fact that "2x" is multiplication, you can take this expression apart and change it into an addition outside the log:
log3(2x) = log3(2) + log3(x)
The answer you are supposed to obtain is:
log3(2) + log3(x)
Do not try to calculate "log3(2)" with your calculator. While you would be correct in saying that "log3(2)" is just a number, they're really asking you to provide the "exact" form of the log, as illustrated above, and not a decimal approximation from your calculator.
Expand log4( 16/x ).
In this example, you have division inside the log, which can be sectioned apart as subtraction outside the log, therefore:
log4( 16/x ) = log4(16) – log4(x)
The first term on the right-hand side of the above equation can be simplified to an exact value, through the application of the basic definition of what a logarithm is:
log4(16) = 2
Then the original expression expands fully as:
log4( 16/x ) = 2 – log4(x)
Constantly remember to take the time to check to see if any of the terms in your expansion (just like the log4(16) above) can be made simpler.
Expand log5(x3)
The exponent inside the log can be brought out front as a multiplier:
log5(x3) = 3 • log5(x) = 3log5(x)
Expand the following:
log2(8x4 ⁄5)
The 5 is divided into the 8x4, therefore divide the numerator and denominator with the use of subtraction:
Log[ 8x4/ 5 ] = Log 2 (8x4) – log2 (5)
Do not take the exponent out front; it is just on the x, not the 8, and you can only take the exponent out front if it is "on" everything inside the log. The 8 is multiplied onto the x4, therefore, take the factors apart with the use of addition:
log2(8x4) – log2(5) = log2(8) + log2(x4) – log2(5)
The x has an exponent (which is now "on" everything inside its log), so therefore move the exponent out front as a multiplier:
log2(8) + log2(x4) – log2(5)
= log2(8) + 4log2(x) – log2(5)
Due to the fact that 8 is a power of 2, I can simplify the first log to an exact value:
log2(8) + 4log2(x) – log2(5)
= 3 + 4log2(x) – log2(5)
Each log is made up of just one thing, therefore this is fully simplified. The answer is:
3 + 4log2(x) – log2(5)
• Expand the following: log3{4(x-5)2 ⁄x4(x-1)3}
Make use of the log rules, and don't try to do too much in one step:
log3{4(x-5)2 ⁄x4(x-1)3} = log3(4(x-5)2)-log3(x4(x-1)3)
= [log3(4)+ log3((x-5)2)-log3(x4)+ log3((x-1)3)]
log3(4)+ log3((x-5)2)-log3(x4)- log3((x-1)3)]
log3(4)+ 2log3(x-5)-4log3(x)+ 3log3(x-1)
Then the final answer is shown as
log3(4) + 2log3(x – 5) – 4log3(x) – 3log3(x – 1)
The logs rules function "backwards", therefore, you can simplify ("compress"?) log expressions. When they tell you to "simplify" a log expression, this normally means they will have given you lots of log terms, each made up of a simple argument, and are requiring you to unite everything into one log with a composite argument. "Simplifying" in this context normally means the opposite of "expanding".
Simplify log2(x) + log2(y).
Since these logs have the same base, the addition outside can be turned into multiplication within:
log2(x) + log2(y) = log2(xy)
The answer = log2(xy).
Simplify log3(4) – log3(5).
Due to the fact that these logs are made up of the same base, the subtraction outside can be turned into division inside:
log3(4) – log3(5) = log3(4/5)
The answer = log3(4/5).
Simplify 2log3(x).
The multiplier out front can be taken within like an exponent:
2log3(x) = log3(x2)
Simplify 3log2(x) – 4log2(x + 3) + log2(y).
You would eliminate the multipliers by transfering them inside as powers:
3log2(x) – 4log2(x + 3) + log2(y)
= log2(x3) – log2((x + 3)4) + log2(y)
Then you'll put the added terms together, and convert the addition to multiplication:
log2(x3) – log2((x + 3)4) + log2(y)
= log2(x3) + log2(y) – log2((x + 3)4)
= log2(x3y) – log2((x + 3)4)
Then I'll account for the subtracted term by linking it inside with division:
log2(x3y) – log2((x + 3)4) = log2 [x3y/(x+3)4 ]
You are supposed to be conversant with these log rules, because there is a particular type of question that the teacher can ask you to test your knowledge of the application of the rules. You cannot get the answer with any type of calculator you may be using.
Question:
Let logb(2) = 0.3869, logb(3) = 0.6131, and logb(5) = 0.8982. With the given values, evaluate logb(10).
Solution
Since 10 = 2 × 5, then:
logb(10) = logb(2 × 5) = logb(2) + logb(5)
Since I have the values for logb(2) and logb(5), I can evaluate:
logb(2) + logb(5) = 0.3869 + 0.8982 = 1.2851
Then logb(10) = 1.2851.
Question:
Let logb(2) = 0.3869, logb(3) = 0.6131, and logb(5) = 0.8982. With the given values, evaluate logb(9).
Since 9 = 32, then:
logb(9) = logb(32) = 2logb(3)
Since I have the value for logb(3), then I can evaluate:
2logb(3) = 2(0.6131) = 1.2262
Then logb(9) = 1.2262.
Let logb(2) = 0.3869, logb (3) = 0.6131, and logb(5) = 0.8982. Using these values, evaluate logb (7.5).
This one is a bit more composite, but, after slightly working with the numbers , you would observe that 7.5 = 15 ÷ 2, thus:
logb(7.5) = logb(15 ÷ 2) = logb(15) – logb(2)
And 15 = 5 × 3, as a result:
logb(15) – logb(2)
= [logb(5) + logb(3)] – logb(2)
= logb(5) + logb(3) – logb(2)
At this point you can compute:
logb(5) + logb(3) – logb(2)
= 0.8982 + 0.6131 – 0.3869
= 1.1244
Therefore, logb(7.5) = 1.1244.
Let logb(2) = 0.3869, logb(3) = 0.6131, and logb(5) = 0.8982. Making use of the given values, evaluate logb(6).
Since 6 = 2 × 3, therefore:
logb(6) = logb(2 × 3) = logb(2) + logb(3)
Because these values have been provided you can evaluate:
logb(2) + logb(3) = 0.3869 + 0.6131 = 1.0000
Therefore, logb(6) = 1.0000.
We obtained that logb(6) = 1. Making use of The Relationship, we would obtain:
logb(6) = 1
b1 = 6
b = 6
Therefore now we know that the base "b" is 6. Although, you may not be given questions and asked to find out the value of the base such as this. There is one more log "rule", although it's more of a formula than a rule. You may have observed that your calculator merely has keys for figuring just the values for the common (base-10) log and the natural (base-e) log, but no other bases.
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