Sure, let's solve the equation step-by-step.
The given equation is:
[tex]$\log_a(2x + 7) = \log_a(x) + 2\log_a(3)$[/tex]
First, let's use the properties of logarithms to simplify the right-hand side of the equation:
1. Recall that [tex]\(2 \log_a(3)\)[/tex] can be rewritten as [tex]\(\log_a(3^2)\)[/tex] using the power rule for logarithms [tex]\((\log_a(b^c) = c \log_a(b))\)[/tex]:
[tex]$\log_a(3^2) = \log_a(9)$[/tex]
Thus, the equation becomes:
[tex]$\log_a(2x + 7) = \log_a(x) + \log_a(9)$[/tex]
2. Next, use the property of logarithms that states [tex]\(\log_a(m) + \log_a(n) = \log_a(mn)\)[/tex]:
[tex]$\log_a(x) + \log_a(9) = \log_a(9x)$[/tex]
So the equation now is:
[tex]$\log_a(2x + 7) = \log_a(9x)$[/tex]
3. Since the logarithms are equal and assuming [tex]\(a > 1\)[/tex], their arguments must be equal:
[tex]$2x + 7 = 9x$[/tex]
4. Solve this linear equation for [tex]\(x\)[/tex]:
First, get all the [tex]\(x\)[/tex] terms on one side of the equation:
[tex]$2x + 7 = 9x$[/tex]
[tex]$7 = 9x - 2x$[/tex]
[tex]$7 = 7x$[/tex]
Divide both sides by 7 to solve for [tex]\(x\)[/tex]:
[tex]$x = 1$[/tex]
Thus, the solution to the equation is:
[tex]$x = 1$[/tex]
