Exercise 1

Solve the exponential equations:

 

1     2^1 - x^2 =\frac{1}{8}

2    \sqrt[3] {8^x} = 655536

3    4^{x^2-6x} = 16384

4    4^\sqrt{x+1} -2 ^\sqrt {x+1}+2 = 0

5    3^{x^2-1} = 134

6    2^{2x} \cdot 2 = 3^x \cdot 3^5

7     3^x \cdot 5^{2x} = 150

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Exercise 2

Solve the following exponential equations.

1     3^ {1-x} -3^x = 2

2     2^{4x} - 2^{2x} -12 = 0

3     e^x - 5e^{-x} + 4e^{-3x} = 0

4     \frac{1}{8} + \frac{1}{4} + \frac{1}{2} + 1 + ....+2^x = \frac {127}{8}

5     4 ^ {x-1} + 2^{x+2} = 48

Exercise 3

Solve the following exponential simultaneous equations.

1      \begin{cases} \frac {2^{2x-3}}{2^{3y+2}} = 2^8 \\ 3x - 2y = 17\end{cases}

2     \begin{cases} 3^x - 2^y = 1 \\ 3^ {x-1} = 2^ {y-2} + 1\end{cases}

3     \begin{cases} 5^x \cdot 25^y = 5^7 \\ 2^ {x-1} \cdot 2 ^ {y+2} = 64\end{cases}

Solution of exercise 1

Now, we will solve the exponential equations from exercise 1 step by step.
1          2 ^ {1 - x^2} = \frac{1}{8}
Since, 2 ^ 3 = 8, hence we can write \frac{1}{8} on the right side of the equation  2 ^ {-3}:

2 ^ {1 - x^2} = 2 ^ {-3}

1 - x^2 = -3

x^2 = 4

x = \pm 2

 

2          \sqrt[3] {8^x} = 655536

We can write the expression \sqrt [3] {8^x} as (2^3) ^ {\frac{x}{3}} = 2^{16}

The resulting expression after simplifying will be:

2^x = 2^{16}

Cancel 2 from both sides of the equation to get the following answer:

x=16

 

3         4^{x^2-6x} = 16384

We can write the above expression as:

2^2 ^ {(x^2 - 6x)} = 2^{14}

Cancel 2 from both sides of the equation to get:

2 ^{x^2} -12x =14

Divide both sides by 2 and set the equation equal to 0 by taking the constant to the right side of the equation:

x^2 -6x -7 = 0

Factor the equation like this:

x^2 -7x + x -7 = 0

x(x - 7) + 1(x - 7) = 0

(x + 1) (x - 7) = 0

Either (x + 1) = 0 or (x - 7) = 0

Hence, x = -1 or x = 7

 

 

4     4^\sqrt{x+1} -2 ^\sqrt {x+1}+2 = 0
Since, 2^2 =4, hence we can write the equation like this:2^{2\sqrt{x+1}} -2 ^\sqrt {x+1}+2 = 0Cancel 2 from both sides of the equation to get the following expression:

2 \sqrt{x + 1} = \sqrt{x + 1} + 2

x = 3

 

5    3^{x^2-1} = 134
Take log on both sides:

log 3 ^ {(x ^ 2 -1)} = log 134

(x ^ 2 - 1)log 3 = log 134

x ^ 2 - 1 = \frac{log134}{log3} = 4.4582

x ^ 2 = 5.4582

x = \pm 2.336

 

6       4 ^ x \cdot 2 = 3 ^x \cdot 243
It can be written as:

\frac{4}{3} =\frac{243}{2}

Take log on both sides of the equation:

log _ {\frac{4}{3}} \frac {4}{3} ^ x = log_{\frac{4}{3}} \frac {243}{2}

x log _ {\frac{4}{3}} \frac {4}{3} = log _ {\frac{4}{3}} \frac {243}{2}

x = \frac{log \frac {243}{2}} { 4} = 16.685

 

7    3^x \cdot 5^{2x} = 150
Take log on both sides of the equation:

log (3^x \cdot 5^2x) = log 150

log 3^x + log 5 ^2x = log 150

x log 3 +2x log 5 = log 150

x (log 3 + 2 log 5) = log 150

log 3 + 2 log 5 =log 3 +log 5^2 = log ( 3 \cdot 5 ^2 ) = log 75

x = \frac {log 150}{log 75} = 1.16

Solution of exercise 2

Solve the following exponential equations.

1       3^ {1-x} -3^x = 2

\frac{3}{3^x} - 3 ^x = 2

Suppose 3 ^ x = t

\frac{3}{t} - t = 2

Solve the expression by taking L.C.M of the left hand side:

\frac{3 - t^2}{t} = 2

Set the equation to 0 by taking the constant to the left hand side of the equation:

\frac{3 - t^2}{t} - 2 = 0

Simplify it further to get the following expression:

t ^ 2 + 2t - 3 = 0

Find factors of the above expression like this:

t ^ 2 + 3t - t - 3 = 0

t(t + 3) -1 (t + 3)=0

(t - 1) = 0 or (t + 3) = 0

t = 1 or t = -3

3^x = 1 or 3^x = -3

The solution for 3^x = -3 cannot be determined, hence the exponential equation will have only one solution, i.e. x = 0

 

2   2^{4x} - 2^{2x} -12 = 0
Put 2 ^ {2x} = t

t^2 - t  - 12 = 0

Solve the above expression by factoring like this:

t^2 -4 t  + 3t - 12 = 0

t (t - 4) + 3(t - 4) = 0

t_1 = 4

t_2 = -3

2 ^ {2x} = 4

2 ^{2x} = -3

2 ^{2x} = -3 has no solution.

2 ^ {2x} = 4 has the solution which is x = 1

 

3   e^x - 5e^{-x} + 4e^{-3x} = 0

We can write the above expression like this:

e ^x - \frac{5}{e ^ x} + \frac {4}{e^{3x}} = 0

e^{4x} - 5e ^{2x} + 4 = 0

Put e ^{2x} = t

t ^2 - 5t + 4 = 0

t_1 = 1     e^{2x} = 1

t_2 = 4    e^{2x} = 4

ln e^{2x} = ln 4

x = 0

x = {ln4}{2}

 

4   \frac{1}{8} + \frac{1}{4} + \frac{1}{2} + 1 + ....+2^x = \frac {127}{8}

\frac{127}{8} = \frac {2 ^x \cdot 2 - \frac{1}{8}} {2 - 1}

\frac{127}{8} = \frac {16 \cdot 2^{x} - 1}{8}

2 ^x = \frac{127}{16}

2^x = 2^3

x = 3

 

4 ^ {x-1} + 2^{x+2} = 48

2 ^ {2x - 2} + 2^ {x+2} = 48

\frac 2^{2x}{4} \cdot 2^x - 48 = 0

Suppose t = 2^x

t ^2 +16t -192 = 0

t ^2 - 24t + 8t -192 = 0

t = 8 or t = -24

8 = 2^x

x = 3

Solution of exercise 3

Solve the exponential simultaneous equations.

1    \begin{cases} \frac {2^{2x-3}}{2^{3y+2}} = 2^8 \\ 3x - 2y = 17\end{cases}

First, we will simplify the first expression and convert it into linear equation in two variables.

Take the expression 2^{3y+2} from the denominator on the left hand side to the right hand side. Hence, the resultant equation will be:

2 ^{2x -3} = 2^8 \cdot 2^{3y + 2}

2x - 3 = 8 +3y + 2

2x - 3y = 13

Now, solve the system of linear equations \begin{cases} 2x - 3y = 13 \\ 3x - 2y = 17\end{cases}

Solving the system of linear equations we get the following values for x and y

x = 5 and y = -1

 

2   \begin{cases} 3^x - 2^y = 1 \\ 3^ {x-1} = 2^ {y-2} + 1\end{cases}

The second expression can be written as:

\frac {3^x}{3} - \frac{2^y}{4} = 1

Put 3^x = u and 2 ^y = v

The resultant system of equations will be:

\begin{cases} u - v = 1 \\ 4u - 3v = 12\end{cases}

Solve the above system of linear equations to get the value of u and v:

u = 9 and v = 8

3^x = 9 and 2 ^y = 8

x = 2 and y = 3

3      \begin{cases} 5^x \cdot 25^y = 5^7 \\ 2^ {x-1} \cdot 2 ^ {y+2} = 64\end{cases}

Since 25 = 5^2 and  64 = 2^6, hence we can write the second expression as:

\begin{cases} 5^x \cdot 5^{2y} = 5^7 \\ 2^ {x-1} \cdot 2 ^ {y+2} = 2^6\end{cases}

Simplifying both the equations further we will get:

\begin{cases} 5^{x + 2y} = 5^7 \\ 2^ {x-1+y+2} = 2^6\end{cases}

\begin{cases} x + 2y = 7 \\ x -1 + y + 2 = 6\end{cases}

x = 3 and y = 2

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Emma

I am passionate about travelling and currently live and work in Paris. I like to spend my time reading, gardening, running, learning languages and exploring new places.

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