Exercise 1

Solve the logarithmic equations.

1   4 log (\frac{x}{5}) + log (\frac{625} {4}) = 2 log x

2    2 log x - 2 log (x + 1) = 0

3    log x = \frac{2 - logx}{log x}

log (25 - x ^ 3) -3 log (4 - x) = 0

5   \frac{log (35 - x ^ 3)}{log (5 - x)} = 3

 

Exercise 2

Solve the logarithmic simultaneous equations.

1           \begin{cases} log x + log y = 2  \\ x - y = 20 \end{cases}

2          \begin{cases} log x + log y = log 2  \\ x^2 + y^2 = 5 \end{cases}

3          \begin {cases} 2 ^ x + 3 ^ y = 31 \\ 2 ^ {x + 1} + 3 ^ {y + 1} = 89\end {cases}

Solution of exercise 1

Solve the logarithmic equations.

4 log (\frac {x}{5}) + log (\frac {625} {4}) = 2 log x

Applying the logarithmic power rule here, we will get the following expression:

log (\frac{x}{5}) ^ 4 +log (\frac{625}{4}) = log x ^ 2

Write the two terms on the left hand side as a single log function by applying logarithm product rule:

log (\frac {x ^ 4 } {625} \cdot \frac{625} {4}) = log x ^ 2

log (\frac {x ^ 4} {4}) = log x ^2

Since both sides of the equation has log functions, so you can write the resultant expression without them like this:

(\frac{x ^ 4} {4}) =  x ^ 2

Set the equation equal to 0 by taking x ^ 2 on the left hand side of the equation:

\frac {x ^ 4} {4 x ^ 2}

The above fraction can be written as:

x ^ 4 - 4 x ^ 2 = 0

Either x ^ 4 = 0 or - 4 x ^2 = 0

Hence, x = 0, x = -2 or x = 2

     
2    2 log x - 2 log (x + 1) = 0
Apply the logarithm power rule to write the above expression as:
log x ^ 2 - log (x + 1) ^ 2 = log 1
Apply the logarithm quotient rule on the left hand side of the equation:
log \frac {x ^ 2} {(x + 1) ^ 2} = log 1
Cancel the log functions from both sides of the equation and solve the resultant equation algebraically:
\frac {x ^ 2} {(x + 1) ^ 2} =  1
Take 2 to the left hand side of the equation:\frac {x ^ 2} {(x + 1) ^ 2} - 1 = 0

x = -\frac {1}{2}

If we substitute x = -\frac{1}{2} in the original equation, we will end up taking the log of negative number which is impossible. Hence, this equation has No Solution.

 

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

By taking the factors from right hand side of the equation to the left hand side and setting the equation to 0, we will get the following expression:

(log x ) ^ 2 + log x - 2 = 0

Suppose log x = t

By substituting the t value in the equation, we will get the following new equation:

t ^ 2 + t - 2 = 0

We will factor the above equation by expanding it and writing the factors in two pairs like this:

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

t( t + 2) - 1(t + 2) = 0

Either (t - 1) = 0 or  (t + 2) = 0

Hence, t = 1 or t = -2

Remember that we assumed log x = t, hence we can say that log x = 1 or log x = -2

By converting the above values in exponential form, we get the following values of x:

x = 10 and x = 10 ^ -2 = \frac {1}{100}

 

4    log (25 - x ^ 3) - 3 log (4 - x) = 0

Apply the power rule here to write the equation as follows:

log (25 - x ^ 3) = log (4 - x ) ^ 3

Cancel the log functions on both sides of the equation to get the following algebraic expression:

(25 - x ^ 3) =  (4 - x ) ^ 3

Use the formula to expand the right hand side of the equation:

25 - x^ 3 = 64 - 48x + 12 x ^2 - x ^3

12 x ^2 -48x +39 = 0

x = 2 \pm \frac {\sqrt{3}}{2}

 

5     \frac{log (35 - x ^ 3)}{log (5 - x)} = 3

Take the expression from the denominator on the left hand side to the numerator on the right hand side of the equation:

log ( 35 - x^3) = 3 log (5 - x)

Apply the logarithm power rule here to get the following equation:

log ( 35 - x ^ 3) = log (5 - x) ^ 3

Cancel the log functions from both sides of the equation and solve the resultant equation algebraically:

( 35 - x ^3) = ( 5 - x)^ 3

x ^ 2 - 5x + 6 = 0

Find factors of above expression by expanding it:

x ^ 2 -3x - 2x + 6 = 0

x (x - 3) -2 (x - 3) = 0

Hence, x = 2 or x = 3

 

Solution of exercise 2

Solve the logarithmic simultaneous equations.

1   \begin {cases} log x + log y = 2 \\ x - y = 20\end {cases}

Use the logarithm product rule on the left hand side of the equation:
log( xy) = 2
In exponential form, it can be written as:
xy = 100
x = \frac{100}{y}
Substitute this value of x in the second equation:
\frac {100}{y} - y = 20
y ^ 2 +20 y - 100 = 0
Use the quadratic formula to find the value of y:y = \frac {-20 \pm \sqrt{400 + 400}}{2}

= \frac {-20 \pm 20 \sqrt {2}}{2}

= -10 + 10 \sqrt {2}

y = 1- (\sqrt {2} - 1)

x = 10 ( \sqrt {2} + 1)

 

2   \begin {cases} log x + log y = log 2 \\ x ^2 + y ^ 2 = 5 \end {cases}

Apply the logarithm product rule on the left hand side of the equation:log (xy) = log 2Cancel the log function from both sides of the equation:
xy = 2
x = \frac {2}{y}
Substitute this value in the second equation:
(\frac {2}{y}) ^ 2 + y ^2 = 5
y ^ 4 - 5y^2 + 4 = 0
y ^ 2 = \frac {5 \pm \sqrt {25 - 16}} {2}y ^ 2 = 4 and y ^ 2 = 1

Hence, y = \pm 2 or y =\pm 1

If y = 2, then x =1

If y = 1 then x = 2

 

3  \begin{cases} 2 ^ x + 3  ^ y =  31 \\ 2 ^ {x + 1} + 3 ^{y + 1} =  89 \end{cases}

We can rewrite the second equation using the exponent product rule:

\begin{cases} 2 ^ x + 3  ^ y =  31 \\ 2 ^ {x} \cdot 2 ^ 1 + 3 ^y \cdot 3 ^ 1 =  89 \end{cases}

Suppose 2 ^ x = m and 3 ^ y = n

\begin{cases} m + n =  31 \\ m \cdot 2 ^ 1 + n \cdot 3 ^ 1 =  89 \end{cases}

\begin{cases} m + n =  31 \\ 2m + 3n =  89 \end{cases}

We will solve this equation through substitution:

m = 31 - n

Substitute this value of m in the second equation:

2 ( 31 - n) + 3n =  89

62 - 2n + 3n = 89

n = 89 -62

n = 27

Put this value of n in the first equation to get the value of m:

m + n = 31

m = 31 - 27

m = 4

Remember that 2 ^ x = m and 3 ^ y = n

Hence, 2 ^ x = 4 and  3 ^ y = 27

Since, 2 raised to the power 2 is equal to 4, so the value of x = 4.

Similarly, 3 raised to the power 3 is equal to 27, so y = 3.

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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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