(1) Use Logarithmic Properties to Expand the Expression

Practice, Test #4, M140

(1) Use logarithmic properties to expand the expression:

骣x2 y ln 琪 (a) 琪 3 桫 z

(2) Use logarithmic properties to write as a single logarithm:

1 ln(x- 1) + ln( x2 + 1) - 3ln( x + 2) 3

(3) Find the derivative of the ln function:

(a) f( x )= x2 ln( x )

(b) g( x )= ln( x2 + x )

(4) Evaluate the integral:

3x2 (a) dx x3 + 5

sec2 x (b) dx tan x

dx (c) 3x - 1

(d) e2x dx

2 (e) xex dx

(5) Solve the exponential equation for k:

6= 12e2460k

(6) Solve the logarithmic equation for x: ln(2x - 3) = 2. (7) Find the derivative of the function:

2 (a) f( x ) = ex-2 x

(b) g( x ) = x2 ex

(c) f(x) = x2ln(x)

(c) f(x) = ln(x2)

(8) Evaluate the integral:

3 (a) 3x2 ex dx

(b) e2x dx

(c) cos(x ) esin x dx

Solutions:

(1) The ln properties are:

(i) ln(xy) = ln(x) + ln(y) (ii) ln(x/y) = ln(x) - ln(y). (iii) r ln x = ln xr

骣x2 y 1 1 1 ln琪 = ln(x2 y2 ) - ln( z 3 ) = ln( x 2 ) + ln( y 2 ) - ln( z 3 ) = 2ln( x ) + ln( y ) - 3ln( z ). 琪 3 桫 z 2

1 1 ln(x- 1) + ln( x2 + 1) - 3ln( x + 2) = ln( x - 1)3 + ln( x 2 + 1) - ln( x + 2) 3 = 3 (2) 3 x-1( x2 + 1) ln (x + 2)3

(3) you have to use the product rule here - and you need to know the rules for differentiating ln functions: d1 d f ( x ) ln(x )= , ln( f ( x )) = . dx x dx f( x ) f( x )= x2 ln( x ), 1 (a) f( x )= 2 x ln( x ) + x2 = x 2x ln( x )+ x .

g( x )= ln( x2 + x ), (b) 2x + 1 g( x ) = x2 + x

Note what happened here - the derivative of the interior function is on top, and the interior function is on the bottom.

3x2 du dx= =ln | u | + C = ln | x3 + 5 | + C . 蝌x3 + 5 u (a) u= x3 +5, ( bigger - deg reed ) du= 3 x2 dx ,

sec2 x du dx= =ln | u | + C = ln | tan( x ) | + C . 蝌tan x u u= tan x du= sec2 x dx ,

(c) dx1 du 1 1 = =ln(u ) + C = ln | 3 x - 1| + C 蝌3x- 1 3 u 3 3 u=(3 x - 1) dx , 1 du= 3 dx ,( rig it : dx :* ) 3 1 du= dx 3 (d) e2x dx u = 2x, du = 2 dx (rig it: dx, ¸ by 2), 1 du= dx Now, substitute in : 2 1 1 e2x dx = eu du= e u + C = (re-sub) 2 2 1 e2x + C. 2 2 (e) xex dx (two polynomials, so: u = x2 (bigger-degreed) du = 2x dx (rig it: need x dx: ¸ 2) 1 du= xdx 2

(5) 6= 12e2460k (Isolate the e part: divide by 12)

6 = e2460k , (ln both sides: ) 12

骣1 2460k ln琪 = lne , 桫2 骣1 ln琪 = 2460k , 桫2 骣1 ln 琪桫2 = k 2460

(6) ln(2x - 3) = 2. (take e to both sides) eln(2x- 3)= e 2 , 2x- 3 = e2 , e2 + 3 x = 2

(7) You need to know the derivative rules for the (natural) exponential function. They are: d d ex= e x, e f( x ) = f ( x ) e f ( x ) dx dx

2 f( x )= ex-2 x ,

(a) 2 f( x )= (2 x - 2) ex-2 x

(b) Here we need to use the product rule: g( x )= x2 ex , g( x )= 2 xex + x2 e x .

(a) Here, we have to use the product rule in conjunction with the the rules for differentiating the logarithmic function ln : d/dx( ln x) = 1/x d/dx( ln (f(x))) = f '(x)/f(x).

(c) f(x) = x2ln(x) g h

f '(x) = 2x ln(x) + x2(1/x) g' h + g h' = 2x ln(x) + x.

(d) f(x) = ln(x2) f '(x) = 2x/x2, f '(x) = 2/x.

(8) The integral rules for the exponential function:

蝌ex dx= e x + C, e u du = e u + C .

The last one is used for u-substitution:

3 3 蝌3x2 ex dx= e u du = e u + C = e x + C (a) u= x3 , ( bigger - deg reed ), du= 3 x2 dx

1 1 e2x dx= e u du = e 2 x + C. 蝌 2 2 u= 2 x , 1 du= 2 dx ,( rig it : dx :* ), 2 1 du= dx, 2

(c) The goal of u-substitution is to pice one part of the integral to be u, and another part to be it's derivative du, here: 蝌cos(x ) esinx dx= e u du = e u + C = e sin( x ) + C u= sin( x ), du= cos( x ) dx