Solving Exponential Equations:
Previously we solved exponential equations where we were able to get a common base
on both sides of the equation.
For example, solve:
32 x  81
Most often it is not going to be obvious what common base to put on both sides, so once the
exponential term is isolated we use the property to take logarithm of base a on both sides
which allows the exponent to be brought out in front of the logarithm as a factor.
For example, solve:
32 x  100
Strategy to solve exponential equations:
1. Isolate an exponential term on one side of the equation
2. Take the logarithm of base a on both sides. (Clearly you
have your choice, but it is often used by taking ln ).
3. Bring the exponent out front as a factor.
4. Solve the resulting equation for the unknown variable & check result.
Solve:
6  51 x   571  653
Exponential_Log Equations blank Page 1
Solving exponential equations continued
Solve:
32 x  100  50
Solve:
Solve:
82 x3  1325
612 x  13x3
Exponential_Log Equations blank Page 2
Factoring and Camouflaged Quadratics
Solve:
Solve:
x 2  3x   49  3 x   0
2e 2 x  e x  15  0
Exponential_Log Equations blank Page 3
Solving Logarithmic Equations
We are trying to find the value of the variable argument that make the equation valid.
• If we have the logarithm of same base on each side of the equation then the argument must
be the same.
• If we have a single logarithm on one side then rewrite it as an exponential to aid in solving for
the unknown variable.
Solve the following for the unknown variable:
(a)
log 6 12  3 x   2
(b)
ln  5 x   8
(c)
4 ln x  1  8
Exponential_Log Equations blank Page 4
Solving Logarithmic Equation continued
Solve the following for the unknown variable:
(d)
log   p  3   log   p  2   log   2 p  24 
(e)
log  x  21  log  x   2
Exponential_Log Equations blank Page 5
Solving logarithmic equations cont.
Solve:
ln  x   ln  x  3   2
Exponential_Log Equations blank Page 6