Exercise 1
Solve the logarithmic equations.
1 
2 
3 
4 
5 
Exercise 2
Solve the logarithmic simultaneous equations.
1 
2 
3 
Solution of exercise 1
Solve the logarithmic equations.
1 
Applying the logarithmic power rule here, we will get the following expression:
Write the two terms on the left hand side as a single log function by applying logarithm product rule:
Since both sides of the equation has log functions, so you can write the resultant expression without them like this:
Set the equation equal to 0 by taking 
on the left hand side of the equation:
The above fraction can be written as:
Either 
or 
Hence, 
, 
or 
If we substitute 
in the original equation, we will end up taking the log of negative number which is impossible. Hence, this equation has No Solution.
3 
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:
Suppose 
By substituting the 
value in the equation, we will get the following new equation:
We will factor the above equation by expanding it and writing the factors in two pairs like this:
Either 
or 
Hence, t = 1 or t = -2
Remember that we assumed , hence we can say that 
or 
By converting the above values in exponential form, we get the following values of 
:
and 
4 
Apply the power rule here to write the equation as follows:
Cancel the log functions on both sides of the equation to get the following algebraic expression:
Use the formula to expand the right hand side of the equation:
5
Take the expression from the denominator on the left hand side to the numerator on the right hand side of the equation:
Apply the logarithm power rule here to get the following equation:
Cancel the log functions from both sides of the equation and solve the resultant equation algebraically:
Find factors of above expression by expanding it:
Hence, or 
Solution of exercise 2
Solve the logarithmic simultaneous equations.
1 
in the second equation:
:
2 
Cancel the log function from both sides of the equation:
and 
Hence, 
or 
If 
, then 
If 
then
3 
We can rewrite the second equation using the exponent product rule:
Suppose 
and 
We will solve this equation through substitution:
Substitute this value of 
in the second equation:
Put this value of 
in the first equation to get the value of :
Remember that and
Hence, 
and 
Since, 2 raised to the power 2 is equal to 4, so the value of 
.
Similarly, 3 raised to the power 3 is equal to 27, so 
.
Find a good maths tutor near me here.
Did you like this article? Rate it!
Loading...
Totally helpful
Some of the equations are not readable (where a dividing liine is used) plus I found one mistake (ln42 instead of ln2)
So I think you should check the page: algebra/ log/ëxponential-equations-worksheet.
Exercise 1.4: I read that as 4^(x-1)^(1/2) – 2(x-1)^(1/2) – 2 = 0. But your solution does not make sense to me.
So probably you mean 4^(x-1)^(1/2) – 2^(x-1)^(1/2) – 2 = 0. But then your solution still does noet make sense to me. If the answer x = 3 then 16 -4 -2 = 10 (not 0).
Exercise 1.1 2^1 – x^2 = 1/8 is not correct. I haven’t looked at the other exercises yet. Please correct your mistakes.