Question

d. \\(\log_{6}(x^{2} - 9) + \log_{6}(3) = 3\\)\\(\log_{4}(10) - \log_{4}(x + 3) = \log_{4}(x)\\)

Explanation:

First Equation: $\boldsymbol{\log_{6}(x^2 - 9) + \log_{6}(3) = 3}$

Step1: Apply log addition rule

Using the property $\log_b(M) + \log_b(N) = \log_b(MN)$, we combine the left - hand side:
$\log_{6}[(x^2 - 9)\times3] = 3$
Simplify the argument: $\log_{6}(3x^2 - 27)=3$

Step2: Convert to exponential form

Recall that if $\log_b(y)=k$, then $y = b^k$. So for $\log_{6}(3x^2 - 27)=3$, we have:
$3x^2 - 27=6^3$

Step3: Calculate $6^3$ and solve for $x$

We know that $6^3 = 216$. So the equation becomes:
$3x^2-27 = 216$
Add 27 to both sides: $3x^2=216 + 27=243$
Divide both sides by 3: $x^2=\frac{243}{3}=81$
Take the square root of both sides: $x=\pm9$

Step4: Check the domain

For the logarithm $\log_{6}(x^2 - 9)$ to be defined, $x^2-9>0$.
When $x = 9$: $9^2-9=81 - 9 = 72>0$
When $x=-9$: $(-9)^2-9 = 81 - 9=72>0$
Both solutions are valid.

Second Equation: $\boldsymbol{\log_{4}(10)-\log_{4}(x + 3)=\log_{4}(x)}$

Step1: Apply log subtraction rule

Using the property $\log_b(M)-\log_b(N)=\log_b(\frac{M}{N})$, we rewrite the left - hand side:
$\log_{4}(\frac{10}{x + 3})=\log_{4}(x)$

Step2: Use the one - to - one property of logs

If $\log_b(M)=\log_b(N)$, then $M = N$ (for $M>0,N>0,b>0,b
eq1$). So we have:
$\frac{10}{x + 3}=x$

Step3: Solve the rational equation

Multiply both sides by $(x + 3)$ (assuming $x
eq - 3$):
$10=x(x + 3)$
Expand the right - hand side: $10=x^2+3x$
Rearrange into standard quadratic form: $x^2+3x - 10 = 0$

Step4: Factor the quadratic equation

We factor $x^2+3x - 10$ as $(x + 5)(x - 2)=0$

Step5: Solve for $x$

Set each factor equal to zero:
$x+5 = 0$ or $x - 2=0$
So $x=-5$ or $x = 2$

Step6: Check the domain

For $\log_{4}(x + 3)$ and $\log_{4}(x)$ to be defined:

• For $\log_{4}(x)$: $x>0$
• For $\log_{4}(x + 3)$: $x+3>0\Rightarrow x>-3$

When $x=-5$: It does not satisfy $x>0$, so we reject it.
When $x = 2$: $x=2>0$ and $x + 3=5>0$, so it is valid.

Answer:

For $\log_{6}(x^2 - 9)+\log_{6}(3)=3$, $x = 9$ or $x=-9$; for $\log_{4}(10)-\log_{4}(x + 3)=\log_{4}(x)$, $x = 2$