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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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

Solve for x.
log 2 x + log 2 (x – 2) = 3
apply the rule of logs: log (MN) = log M + log N
so log 2 x (x – 2) = 3
Now we change to exponential form
2 3 = x (x – 2) = 8 since 2 3 = 8
x 2 – 2x – 8 = 0 becomes (x – 4)(x + 2) = 0 so x = 4 or x = –2 .

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  .

Sometimes, we have to solve for an unknown base, like this

Example 5  .

| 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.

a) Nick buys a car for \$25,000. The depreciation rate for the first year is 30%
and it is 15% for every subsequent year he owns the car.
i) How much will the car be worth at the end of the first year?
ii) If Nick intends to sell the car when its value falls below \$10,000
how many years after purchase does he sell the car?

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b) Olivier gets \$3000 as a gift for his birthday.
He invests it at 7.5 % compounded yearly.
Lauren too gets a gift of \$4000 which she invests at 7% compounded yearly.
They decide they will stay friends until they have the same amount in their accounts.
For how long will Olivier and Lauren be friends?

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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 =) – 2x – 6 = 4 so x = – 5 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 = 4log 3 2x = 4 becomes 3 4 = 2x x = 81 / 2 b) log 6 (x + 3) + log 6 (x – 2) = 1log6 [(x + 3)(x – 2)] = 1 x 2 + x – 12 = 0 so x = 3 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 – 5log x (1 / 9) – log x 27 = – 5, (log x 1 = 0) so x = 3

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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

x + y = 38

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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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