Transcript Solving Exponential and Logarithmic Equations

Solving Exponential and
Logarithmic Equations
Exponential Equations are equations of the
form y = abx. When solving, we might be
looking for the x-value, the b value or the
y- value. First, we’ll review algebraic methods.
When solving for b, isolate the b value; then
raise both sides of the equation to the
reciprocal power of the exponent.
54  2b
3
27  b
3
 27 
1
3

1
3 3
b
 
3b
When solving for y, solve by
performing the indicated operations.
y  16 
1

2
1
2
 1
y 
 16 
1
y
4
When solving for the exponent,
rewrite the bases so they have the
same base. If the bases are equal, the
exponents are equal. Now, solve.
8 4
x
2 x 1
3
2
2
x
x1
 
3 x  2( x  1)
3x  2 x  2
x2
Both bases now equal 2
so we can just use the
equal exponents.
Solve the following exponential
equations:
81  3b
3 = b
3
y  125 
2
3
y = 25
1 x
1
 
3
9
x
Solving Exponential Equations by Using
the Graphing Calculator
•
•
•
•
Always isolate the variable FIRST!!
Graph the function in Y1
Graph the rest of the equation in Y2
Use the intersect function (found
with
) to determine the
x value
Solve for x to the nearest thousandth:
ex=72
• Graph Y1 = ex and Y2= 72
• Use intersect to give the answer
x =4.277
Log Equations are of the form y=logba or
y= ln x where the base is e.
Y= log216
4=log3x
3= logx1000
To solve for x, we need to undo the log format
by rewriting in exponential form.
2y=16
34=x
x3=1000
Now we use the exponential rules to solve.
2y=16
y=3
81 = x
X3 = 103
x=4
Solve algebraically:
Rewrite as an ln equation:
e  72
x
ln e  ln 72
x
Since ln ex means the exponent of ln ex, just use x:
x  ln 72
x  4.277
Note: when an equation is written in
terms of e, you MUST use natural logs.
Otherwise you may use log or ln at will.
2x = 14
Solve for x algebraically:
Take the ln or log of each side and solve.
ln 2  ln14
x ln 2  ln14
ln14
x
ln 2
x  3.807
x
log 2  log14
x
x log 2  log14
log14
x
log 2
x  3.807
Solve for x: 2x=14 graphically.
• Graph Y1 = 2 x and Y2= 14
• Use intersect to approximate the answer
X=3.807
Solve lnx = 3 to the nearest tenth.
ln x  3
e3  x
X = 20.1
Solve for x:
2  32
x
X= 5
x
1
  9
3
X=-2
More Involved Equations
Sometimes log or ln equations require a few
more steps to “clean them up” before we can
simply “take the log” or “undo the log”.
5  2ln x  4
2ln x = -1
ln x = -0.5
e-0.5=x
x= 0.60653…
Before you can take the ln,
you need to isolate it!
To solve graphically, you still must
isolate the ln expression.
2ln x = -1
Graph as y1 = 2lnx
y2= -1
Solve ln 3x  2
e 2  3 x  definition of log
x  2.46
Solve 2log5 3x  4
log 5 3 x  2
52  3 x  log definition
25
x
 8.3333
3
Solve using the graphing calculator:
ln x  x 2  2
ln x  x2  2  0
More practice
3(2 )  42
x
83x  360
log 2 x  log14
x log 2  log14
log14
x
 3.807354922
log 2
83 x  360
ln 83 x  ln 360
3 x ln 8  ln 360
ln 360
3x 
ln 8
ln 360
x
3 ln 8
x  .93453023
Solving Logarithmic Equations
Algebraically Using Laws of Logarithms
When an equation contains the word log or ln, we need
to eliminate it to solve the equation so first we apply
the laws of logarithms to “undo” the addition by
changing to multiplication, “undo” subtraction by
changing it to division, and “undo” powers by changing
them to multiplication..
Solve:
Log 2(4x+10) – log2(x+1) = 3
Log 2(4x+10) – log2(x+1) = 3
Apply Quotient Rule.
4 x  10
log2
3
x 1
4 x  10
 23
x 1
4 x  10  8( x  1)
4 x  10  8 x  8
2  4x
1
x
2
Definition of Logarithm
Cross multiply and solve
Solve 4e  3  2
2x
4e 2 x  5
5
2x
e 
4
5
ln e  ln
4
5
2 x  ln
4
1 5
x  ln  0.11
2 4
2x
Solve:
5 7
3x
x1
Solve e  3e  2  0
2x
x
2t5
Solve 2(3
)  4  11
2(32t 5 )  4  11
2(32t 5 )  15
2 t 5
3
15

2
log 3 (32t 5 )  log 3
15
2
15
2t  5  log 3  Inverse Property
2
5
15
t=  log 3  3.42
2
2