Solving Exponential Equations
Determine the Value of X that makes the equation true (or
satisfies the equation)
8^{ x  2} = 2^{ x + 4}
Solution 1 guess and check
Sub in different values of x, trying to get closer and closer.
X 
LS 
RS 
LS = RS 
1 
1/8 
32 
NO 
2 
1 
64 
NO 
5 
512 
512 
YES 
Not a very efficient way of solving equations.
Solution 2 equating the bases
If an equation can be rearranged so that the bases are the
same, this means the exponents than have to be equivalent as a
result.
In this case we can rewrite 8 with a base of 2 and an
exponent of 3. Using exponent laws we see that:
8^{ x  2} 
= 2^{ x + 4} 
( 2^{ 3 }) ^{x 
2} 
= 2^{ x + 4} 
2^{ 3x  6 } 
= 2^{ x + 4} 
3x  6 
= x + 4 
2x 
= 10 
x 
= 5 
Other methods:
Â· Get the bases to be the same
Â· Factor out a common factor (involving an exponent)
3^{ x + 2}  3^{
x } 
= 216 

3^{ x }( 3^{ 2}
 1) 
= 216 

3^{ x }( 9  1) 
= 216 
find a common factor by
remembering your exponent laws (a^{m} a^{n}
= a^{m+n} ) and then follow BEDMAS rules. Then
find a common base and solve for x. 
3^{ x } 

3^{ x } 
= 27 
3^{ x } 
= 3^{ 3} 

x 
= 3 

Â· Note knowing the powers of 2 from 0 to 10 is very helpful,
same with the powers of 3 from 1 to 5.
REMEMBER
Â· Reduce the expression to only two parts (one with an
unknown exponent and a constant term)
o Common methods Factoring, BEDMAS
rules, Exponent laws
Â· Find a common base
o NOTE zero can be an base with an exponent zero.
