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 =)
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so x = 5 |
(divide by 4) 3x / 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 log6 [(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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