Question

question 5/24
5
a) solve the following equation.
\log_{4}(2x + 5) + \log_{4}(3) = \log_{4}(3x)
x =

b) the solution is , because the argument is . therefore,
it is a solution
there is no solution

Explanation:

Step1: Apply log addition rule

Using the property \(\log_b(M) + \log_b(N)=\log_b(MN)\), we rewrite the left - hand side of the equation \(\log_4(2x + 5)+\log_4(3)=\log_4(3x)\) as \(\log_4[3(2x + 5)]=\log_4(3x)\).

Step2: Use one - to - one property of logs

If \(\log_b(a)=\log_b(c)\), then \(a = c\) (for \(b>0,b
eq1,a>0,c>0\)). So we have the equation \(3(2x + 5)=3x\).

Step3: Solve the linear equation

First, expand the left - hand side: \(6x+15 = 3x\).
Subtract \(6x\) from both sides: \(15=3x - 6x=-3x\).
Divide both sides by \(- 3\): \(x=-5\).

Step4: Check the domain of the logarithm

For a logarithmic function \(\log_b(u)\), \(u>0\).

• For \(\log_4(2x + 5)\), when \(x = - 5\), \(2x+5=2\times(-5)+5=-10 + 5=-5<0\).
• For \(\log_4(3x)\), when \(x=-5\), \(3x=3\times(-5)=-15<0\).

Since the arguments of the logarithms are negative when \(x = - 5\), \(x=-5\) is not in the domain of the original logarithmic equation.

Answer:

a) \(x=-5\) (but we will see in part b that this is not a valid solution)
b) The solution is \(-5\), because the argument is negative. Therefore, there is no solution