# Tutor profile: Thu N.

## Questions

### Subject: Pre-Calculus

expand logarithm: $$ \log_4(\frac{2^2}{3})^3$$

$$ \log_4(\frac{2^2}{3})^3$$ = $$3 \log_4(\frac{2^2}{3})$$ = $$ 3(\log_42^2 - \log_43) $$ = $$ 3\log_44 - 3\log_43 $$ = $$ 3 - 3\log_43 $$

### Subject: Pre-Algebra

solve for x and y: $$ 5x^{2} - 3y = -15 $$ $$ 2x + y = 5 $$

$$ 5x^{2} - 3y = -15 $$ (1) $$ 2x + y = 5 $$ (2) (1) $$ -> y = 5 - 2x $$ so (2) : $$ 5x^{2} - 3*(5 - 2x) = -15 $$ $$ 5x^{2} - 15 + 6x = -15 $$ $$ 5x^{2} + 6x = -15 + 15 $$ $$ 5x^{2} + 6x = 0 $$ $$ x(5x + 6) = 0$$ $$ -> x = 0$$ or $$5x + 6 = 0$$ $$ x = 0$$ or $$ x = -6/5$$

### Subject: Algebra

Simplify to lowest terms: $$ \frac{x^8+a^3 * x^4 * y}{x^8-a^6 * y^2 } $$

$$ \frac{x^8+a^3 * x^4 * y}{x^8-a^6 * y^2 } $$ = $$ \frac{x^4*(x^4 + a^3 * y)}{(x^4)^2 - (a^3 * y)^2 } $$ = $$ \frac{x^4*(x^4 + a^3 * y)}{(x^4 + a^3 * y)* (x^4 - a^3 * y) } $$ = $$ \frac{x^4}{(x^4 - a^3 * y) } $$

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