$$ imes $$ One or • more of your responses is incorrect. First, use the rule that $$ \log _{\mathrm{a}} \mathrm{M}^{\mathrm{r}}=\mathrm{r} \log _{\mathrm{a}} \mathrm{M} $$. Then, use the product rule for logarithms, which states that for any positive real numbers M and N and any positive base $$ \mathrm{b}
eq 1, \log _{\mathrm{b}} \mathrm{MN}=\log _{\mathrm{b}} \mathrm{M}+\log _{\mathrm{b}} \mathrm{N} $$. Finally, solve for x . Check your proposed solution(s) in the original equation. Be sure to reject any values that produce a logarithm of a negative number or zero.

Question

$$ 	imes $$ One or • more of your responses is incorrect. First, use the rule that $$ \log _{\mathrm{a}} \mathrm{M}^{\mathrm{r}}=\mathrm{r} \log _{\mathrm{a}} \mathrm{M} $$. Then, use the product rule for logarithms, which states that for any positive real numbers M and N and any positive base $$ \mathrm{b} 
eq 1, \log _{\mathrm{b}} \mathrm{MN}=\log _{\mathrm{b}} \mathrm{M}+\log _{\mathrm{b}} \mathrm{N} $$. Finally, solve for x . Check your proposed solution(s) in the original equation. Be sure to reject any values that produce a logarithm of a negative number or zero.

UpStudy AI · Accepted Answer

To solve a logarithmic equation using the properties you've mentioned, let's consider a general example. Suppose we have the equation:

$$
\log_a(M^r) = \log_a(N)
$$

We can apply the power rule of logarithms:

$$
r \log_a(M) = \log_a(N)
$$

Next, we can isolate $$\log_a(M)$$:

$$
\log_a(M) = \frac{\log_a(N)}{r}
$$

Now, if we want to solve for $$x$$ in a more specific equation, let's say:

$$
\log_a(2x) + \log_a(3) = 2
$$

We can use the product rule for logarithms:

$$
\log_a(2x \cdot 3) = 2
$$

This simplifies to:

$$
\log_a(6x) = 2
$$

Now, we can rewrite this in exponential form:

$$
6x = a^2
$$

Next, we solve for $$x$$:

$$
x = \frac{a^2}{6}
$$

Now, we need to check if this solution is valid. We must ensure that $$2x$$ and $$3$$ are both positive, as logarithms are only defined for positive arguments.

1. **Check $$2x > 0$$**:
   $$
   2x = 2 \cdot \frac{a^2}{6} = \frac{a^2}{3}
   $$
   This is positive as long as $$a > 0$$.

2. **Check $$3 > 0$$**:
   This is always true.

Thus, the solution $$x = \frac{a^2}{6}$$ is valid as long as $$a > 0$$.

If you have a specific equation in mind, please provide it, and I can help you solve it step by step!
To solve a logarithmic equation, apply the power rule and product rule of logarithms. Solve for $$x$$ and ensure the arguments of the logarithms are positive.

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