$\bf \textit{Logarithm of rationals} \\\\ \log_a\left( \frac{x}{y}\right)\implies \log_a(x)-\log_a(y) \\\\[-0.35em] \rule{34em}{0.25pt}\\[1em] [\log(3)-\log(4)]-\log(2)\implies \log\left( \cfrac{3}{4} \right)-\log(2)\implies \log\left( \cfrac{\frac{3}{4}}{~~2~~} \right) \\\\\\ \log\left( \cfrac{\frac{3}{4}}{~~\frac{2}{1}~~} \right)\implies \log\left( \cfrac{3}{4}\cdot \cfrac{1}{2} \right)\implies \log\left( \cfrac{3}{8} \right)$

4 0

3 years ago

Answer, please. I need help.

Ghella [55]

Answer:

-8x^12

Step-by-step explanation:

When you have an exponent outside of parenthesis, it is raising everything in the parenthesis to that exponent (in this case, the third). The first step to solve this is to multiply the exponents in the parenthesis by 3. This is called the power of a power rule.

-2 is being raised to the first power. 1x3 = 3, and x is being raised to the fourth power. 4x3 = 12, so you now have:

-2^3x^12

You can simplify -2^3: -2 x -2 x -2 = -8

This leaves you with -8x^12

You can check this by putting -8x^12 and (-2x^4)^3 into a graphing calculator. You know that you simplified correctly if they show the same graphs.

7 0

3 years ago