Exponents and Logarithms Quiz

1. Log of a Product:

   $$\log\left(x\cdot y\right)=\log x+\log y$$ because $$b^x\cdot b^y=b^{x+y}$$

2. Log of a Quotient:
   $$\log\left(\frac{x}{y}\right)=\log x-\log y$$

   because $$\frac{b^x}{b^y}=b^{x-y}$$

3. Log of an Exponent:

   $$\log x^a=a\cdot\log x$$ because $$\left(b^x\right)^y=b^{x\cdot y}$$

4. Change of the Base:

   $$\log_bx=\frac{\log x}{\log b}$$ because $$b^x=a^{\log_bb^x}$$

5. Log of 1:
   $$\log\left(1\right)=0$$ because $$b^0=1$$

$$\log_bb^x=x$$ , and $$b^{\log_bx}=x$$

So logarithms and exponents are inverse operations, in the same sense that multiplication and division by the same number are inverse operations (They cancel each other out).

Steps to solve:

1. Take log base 3 of both sides.

2. On the left, apply the "log of a product" property.

3. On the left, apply the "log of an exponent" property.

4. On the left, apply the "log of the base" property.

5. Move all three terms to the same side to get:

   $$x^2+\left(\log_32\right)\cdot x-\log_36=0$$

6. Substitute $$a=1,b=\log_32$$ , and $$c=-\log_36$$ into the quadratic formula and evaluate to find $$x$$ .

$$2\cdot\left(2^{x-1}-1\right)=2^x-2$$
If we look at the end behaviour:
$$\lim_{x\to-\infty}\left(2^x-2\right)$$

We find that $$2^x\to0$$ . Then $$2^x-2\to-2$$ .

The equation has 2 real solutions. Solve by taking $$\log_4$$ of both sides and applying the logarithm properties. Then, once you arrive at a quadratic equation, use the quadratic formula to solve.

$$2^x\cdot4^{x^2}=8$$

$$\log_4\left(2^x\cdot4^{x^2}\right)=\log_48$$

$$x\cdot\log_42+x^2=\log_48$$

$$x^2+\left(\log_42\right)\cdot x-\log_48=0$$

Since the range of $$y=b^x$$ is $$\left\{x|\ x>0\right\}$$ for any base, then the domain of $$y=\log_bx$$ must be $$\left\{x|\ x>0\right\}$$ for any base.