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( x 3 5) 8 = 9
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We have ( x 3 5) 8 = 9 x 3 5 = 91/8 x 3 = 91/8 5 1 x = ( 91/8 5 1 ) 1/3 x = 91/24 51/3
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YOU TRY IT Solve the equation 4x 32x = 7 [Hint: Take the logarithm of both
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If you review the first few paragraphs of Section 1, you will find an intuitively appealing definition of the logarithm to the base 2: log2 x is the power to which you need to raise 2 to obtain x With this intuitive notion we readily see that log2 16 = the power to which we raise 2 to obtain 16 = 4 and log2 (1/4) = the power to which we raise 2 to obtain 1/4 = 2 However, this intuitive approach does not work so well if we want to take log 5 or log2 7 Therefore we will give a new definition of the logarithm to any base a > 0 which in simple cases coincides with the intuitive notion of logarithm If a > 0 and b > 0 then loga b = ln b ln a
EXAMPLE
Calculate log2 32
SOLUTION
We see that log2 32 = ln 25 5 ln 2 ln 32 = = = 5 ln 2 ln 2 ln 2
Notice that, in this example, the new definition of log2 32 agrees with the intuitive notion just discussed
CALCULUS DeMYSTiFieD
EXAMPLE
Express ln x as the logarithm to some base
SOLUTION
If x > 0 then loge x = ln x ln x = = ln x ln e 1
Thus we see that the natural logarithm ln x is precisely the same as loge x
MATH NOTE In mathematics, it is common to write ln x rather than loge x YOU TRY IT Calculate log3 27 + log5 ( 1/25) log2 8
We will be able to do calculations much more easily if we learn some simple properties of logarithms and exponentials If a > 0 and b > 0 then a(loga b) = b If a > 0 and b R is arbitrary then loga (ab ) = b If a > 0, b > 0, and c > 0 then (i) loga (b c) = loga b + loga c (ii) loga (b/c) = loga b loga c (iii) loga b = (iv) loga b = logc b logc a 1 logb a
(v) loga 1 = 0 (vi) loga a = 1 (vii) For any exponent , loga (b ) = (loga b)
6 T R A N S C E N D E N T A L F U N C T I O N S
We next give several examples to familiarize you with logarithmic and exponential operations
EXAMPLE
log3 81 5 log2 8 3 ln( e4 )
Simplify the expression
SOLUTION
The expression equals log3 ( 34 ) 5 log2 ( 23 ) 3 ln e4 = 4 log3 3 5 3 log2 2 3 4 ln e = 4 1 5 3 1 3 4 1 = 23
YOU TRY IT What does log3 5 mean in terms of natural logarithms
EXAMPLE
5x 23x = 4 7x
Solve the equation
for the unknown x
SOLUTION
We take the natural logarithm of both sides: ln( 5x 23x ) = ln 4 7x
Applying the rules for logarithms we obtain ln( 5x ) + ln( 23x ) = ln 4 ln( 7x ) or x ln 5 + 3x ln 2 = ln 4 x ln 7
CALCULUS DeMYSTiFieD
Gathering together all the terms involving x yields x [ln 5 + 3 ln 2 + ln 7] = ln 4 or x [ln( 5 23 7) ] = ln 4 Solving for x gives x= ln 4 = log280 4 ln 280
EXAMPLE
B= 5 log7 3 ( 1/4) log7 16 3 log7 5 + ( 1/5) log7 32
Simplify the expression
SOLUTION
The numerator of B equals log7 ( 35 ) log7 ( 161/4 ) = log7 243 log7 2 = log7 ( 243/2) Similarly, the denominator can be rewritten as log7 53 + log7 ( 321/5 ) = log7 125 + log7 2 = log7 ( 125 2) = log7 250 Putting these two results together, we find that B= log7 243/2 log7 250
= log250 ( 243/2)
YOU TRY IT What does 3
tion) EXAMPLE
mean (in terms of the natural logarithm func-
Simplify the expression ( log4 9) ( log9 16)
6 T R A N S C E N D E N T A L F U N C T I O N S
SOLUTION
We have ( log4 9) ( log9 15) = 1 log9 4 log9 16
= log4 16 = 2
64 Calculus with Logs and Exponentials to Arbitrary Bases 641 Differentiation and Integration of loga x and a x
We begin by noting these facts: If a > 0 then (i) (ii)
d x a dx
= ax ln a; equivalently, x) =
1 x ln a
a x dx =
ax ln a
+ C
d (loga dx
Still Struggling
As always, we can state these last formulas more generally as du d u a = au ln a dx dx and d 1 du 1 loga u = dx u dx ln a