Question 4 Solve the logarithmic equations (YOU MUST SHOW ALL WORK FOR FULL CREDIT). a) x=log_(5)320 4.(log_(3)(1)/(27))=x c) log_(3)(5x-12)=2 d) ln((x)/(4)+7)=3 e) log_(x+3)(49)=2 e) log_(2x-1)(625)=4 (10 point)

Let's solve each logarithmic equation step by step.

### a) $x = \log_{5}320$

To solve this, we need to express 320 as a power of 5:
$$320 = 5^x$$

Taking the logarithm base 5 of both sides:
$$x = \log_{5}320$$

This equation does not have a simple solution in terms of elementary functions, so we leave it as:
$$x = \log_{5}320$$

### b) $4 \cdot \log_{3}\left(\frac{1}{27}\right) = x$

First, simplify the logarithm:
$$\log_{3}\left(\frac{1}{27}\right) = \log_{3}(3^{-3}) = -3$$

Now multiply by 4:
$$4 \cdot (-3) = x$$
$$x = -12$$

### c) $\log_{3}(5x - 12) = 2$

Rewrite the equation in exponential form:
$$3^2 = 5x - 12$$
$$9 = 5x - 12$$

Solve for $x$:
$$5x = 21$$
$$x = \frac{21}{5}$$
$$x = 4.2$$

### d) $\ln\left(\frac{x}{4} + 7\right) = 3$

Rewrite the equation in exponential form:
$$e^3 = \frac{x}{4} + 7$$

Solve for $x$:
$$\frac{x}{4} = e^3 - 7$$
$$x = 4(e^3 - 7)$$

Using $e^3 \approx 20.0855$:
$$x = 4(20.0855 - 7)$$
$$x = 4 \cdot 13.0855$$
$$x \approx 52.343$$

### e) $\log_{x+3}(49) = 2$

Rewrite the equation in exponential form:
$$(x+3)^2 = 49$$

Solve for $x$:
$$x+3 = \pm 7$$

So,
$$x + 3 = 7 \quad \text{or} \quad x + 3 = -7$$
$$x = 4 \quad \text{or} \quad x = -10$$

Since logarithms are only defined for positive arguments, we must check the domain:
$$x + 3 > 0$$
$$x > -3$$

Thus, the valid solution is:
$$x = 4$$

### f) $\log_{2x-1}(625) = 4$

Rewrite the equation in exponential form:
$$(2x-1)^4 = 625$$

Take the fourth root of both sides:
$$2x - 1 = \sqrt[4]{625}$$
$$2x - 1 = 5$$

Solve for $x$:
$$2x = 6$$
$$x = 3$$

### Summary of Solutions:
a) $x = \log_{5}320$
b) $x = -12$
c) $x = 4.2$
d) $x \approx 52.343$
e) $x = 4$
f) $x = 3$