 Section 5-7:  Change of Base Formula Try the quiz at the bottom of the page!
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An exponential equation is an equation that contains a variable in the exponent.  We solved problems of this type in a previous chapter by putting the problem into the same base.  Unfortunately, it is not always possible to do this.  Take for example, the equation 2x = 17.  We cannot put this equation in the same base.  So, how do we solve the problem?  We use the change of base formula!!  We can change any base to a different base any time we want.  The most used bases are obviously base 10 and base e because they are the only bases that appear on your calculator!! Change of base formula Logb x = Loga x/Loga b
Pick a new base and the formula says it is equal to the log of the number in the new base divided by the log of the old base in the new base.

Examples

1)  Log2 37 =
Solution:  Change to base 10 and use your calculator.
=  Log 37/log 2
Now use your calculator and round to hundredths.
= 5.21
This seems reasonable, as the log2 32 = 5 and log2 64 = 6.

2)  Log7 99 =
Solution:  Change to either base 10 or base e.  Both will give you the same answer.  Try it both ways and see.
= Log 99/Log 7    or  Ln 99/ Ln 7
Use your calculator on both of the above and prove to yourself that you get the same answer.  Both ways give you 2.36.

Solving Exponential Equations using change of base

Now, let's go back up and try the original equation:
2x = 17
To put these in the same base, take the log of both sides.  Either in base 10 or base e.  Hint.  Use base e only if the problem contains e.
Log 2x = Log 17
Using the log rules, we can write as:
x Log 2 = Log 17
Now isolate for x and use your calculator.
x = Log 17/Log 2
x = 4.09
To check your answer, type in 24.09 and see what you get!  The answer will come out slightly larger than 17 do to rounding.

Sample Problems

1)  e3x = 23
Solution:  Use natural log this time.
Ln e3x = Ln 23
3x Ln e = Ln 23
3x = Ln 23  ( Ln e = 1)
x = (Ln 23)/3
x = 1.05

2)  How long does it take \$100 to become \$1000 if invested at 10% compounded quarterly?
Solution:  Ao = 100, A(t) = 1000, r = .1,  n = 4
1000 = 100(1 + .1/4)4t
10 = 1.0254t   Use the change of base formula
Log 10 = Log 1.0254t
1 = 4t Log 1.025  (Log 10 = 1)
1/(4Log 1.025) = t
t = 23.31
It will take 23.3 years to have \$1000 from the \$100
investment. I am so ready for the test!  Bring it on!! I'm a little sick to my stomach.  Better back up!   Current quizaroo #  5b 1)   Find x if  log(x2 -21) = 2

a)  11
b)  5, -5
c)  11, -11
d)  0
e)  no solution 2)  Expand the log:         log(x/yz)

a) log x + log y - log z

b) log x - log y + log z
c) log x - (log y - log z)
d) log x - log y - log z
e) log x + log (yz) 3)  Find the best solution for    log 5 + log 40 - log 2

a)  log 100
b)  200
c)  log 200
d)  log 38
e)  2 4)  Find the value of   log4 95 to thousandths place.

a) .304
b) 3.285
c) 1.978
d) 4.554
e) .494 5)  Solve the equation:  43x-1 = 3    Round to the nearest thousandth.

a) 1.792
b) 1.126
c) .597
d) .793
e) 1.333  