Solving Exponential and Log Equations 
Solving Exponential Equations
If you can express everything as powers of the same base,
do so, then set the powers equal and solve.
example 1
so, x + 2 = 4x – 7 and x = 3
If both sides cannot be expressed as powers of the same base,
take log or ln of both sides and solve.
memory tweak: log (m)^{ n} = n log m, therefore log (x)^{ 6} = 6 log x
example 2
2^{ 2 x + 3} = 5^{ x} – 4 becomes (2x + 3) log 2 = (x – 4) log 5
2x log 2 + 3 log 2 = x log 5 – 4 log 5
x (2 log 2 – log 5) = –3 log 2 – 4 log 5
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 exponential equations  log equations  practice  solutions 
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Solving Log Equations
memory tweak: since the base of any log function is positive (> 0),
we can't take the log of negative numbers.
Solutions which make the log function argument negative must be dumped.
Example 3
We can't take log_{ 2} (–2) since there is no power of 2 that gives us –2. The solution is x = 4.
To solve a log equation, we express the terms as a single log = a constant,
change to exponential form, then solve for the unknown.
Note: watch for logs you can evaluate before solving like this
Example 4
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Sometimes, we have to solve for an unknown base, like this
Example 5
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 exponential equations  log equations  practice  solutions 
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1/ Solve these exponential equations
a) 5^{ – 2 (x + 3)} = 625  
d) 2^{ x }– 2 = 32^{ x} + 2 .  
e) 3^{ 4x – 1} = 15  f) 3(0.5)^{5 – 2x} = 21 
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2/ Solve these log equations. Identify any restrictions.
a) log_{ 3} x + log_{ 3} 2 = 4  b) log_{ 6} (x + 3) + log_{ 6} (x – 2) = 1 
c) 3 log (0.5x + 2)^{ 2} – 1 = 5  d) log_{ x} (1 / 9) – log_{ }_{x} 1 = log_{ }_{x} 27 – 5 
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3/ Given 2^{ y} = 16^{ x }– 3 and 3^{ y} + 2 = 27^{ x} , find the value of x + y.
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4/ Solve these word problems.
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 exponential equations  log equations  practice  solutions 
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1/ Solve these exponential equations
a) 5^{ – 2 (x + 3)} = 625 = 5^{ 4} . (set exponents =)

so x = 5 
(divide by 4) 3^{x }/ 2 = 27 = 3^{ 3}, therefore ½(x) = 3, so x = 6 
d) 2^{ x – 2 } = 32^{ x }+ 2 = 2^{ 5(x + 2)} , x – 2 = 5x + 10 so x = – 3 
e) 3^{ 4x – 1} = 15 (no common base here) (4x – 1) log 3 = log 15
x = 0.8662 
f) 3(0.5)^{5 – 2x} = 21 (divide by 3) (0.5)^{5 – 2x} = 7 so (5 – 2x) log(0.5) = log 7 solving for x we get 3.90 
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2/ Solve these log equations. Identify any restrictions.
a) log_{ 3} x + log_{ 3} 2 = 4 log_{ 3} 2x = 4 becomes 3^{ 4} = 2x x = 81 / 2 
b) log_{ 6} (x + 3) + log_{ 6} (x – 2) = 1 log_{6} [(x + 3)(x – 2)] = 1 
c) 3 log (0.5x + 2)^{ 2} – 1 = 5 log (0.5x + 2)^{ 2} = 2 (0.5x + 2)^{ 2} = 10^{ 2}, so 0.5x + 2 = 10 so x = 16 
d) log_{ x} (1 / 9) – log_{ }_{x} 1 = log_{ }_{x} 27 – 5 log_{ x} (1 / 9) – log_{ }_{x} 27 = – 5, (log_{ x} 1 = 0)

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3/ Given 2^{ y} = 16^{ x }– 3 and 3^{ y} + 2 = 27^{ x} , find the value of x + y.
Since 2^{ y} = (2^{ 4}) ^{( x – 3) } and 3^{ y} + 2 = (3^{ 3})^{ x},
When we set the exponents equal to each other we get
y = 4x – 12 and y = 3x – 2
So 4x – 12 = 3x – 2, so x = 10 and y = 28
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4/ a) after 1st year, the car is worth 0.7($25,000) = $17,500
so V(t) = 17,500(0.85)^{ t}
we want V(t) < 10,000 so we set 17,500(0.85)^{ t} < 10,000
The car will be worth less than $10,000 after 4.44 years.
b) Olivier: $3000(1.075)^{ t}  Lauren: $4000(1.07)^{ t} . 
$3000(1.075)^{ t} = $4000(1.07)^{ t}
log 3 + t (log 1.075) = log 4 + t (log 1.07)
t (log 1.075 – log 1.07) = log 4 – log 3
t = = 61.71 years
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 exponential equations  log equations  practice  solutions 
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