Solve the following math problems: 1. Given that 2^x / 4^y = 1 and 3^x * (sqrt(3))^y = 9, obtain two linear equations in x and y. Hence determine the values of x and y. 2. Solve th... Solve the following math problems: 1. Given that 2^x / 4^y = 1 and 3^x * (sqrt(3))^y = 9, obtain two linear equations in x and y. Hence determine the values of x and y. 2. Solve the equation (e^(3x) - e^(-3x)) / 2 = 4, giving your solution correct to 3 s.f. 3. By using substitution y = e^x, solve the equation 2e^(2x) - 5e^x + 2 = 0, leaving solutions in exact form. 4. Solve log_10(2x^3 + 1) - 3log_10(x) = 1 5. Solve the simultaneous equations: 2^x + 3^y = 5 and 2^(x+2) - 3^(y+1) = 13 6. Express ln((2x^2 - 3x - 2) / sqrt(1-x)) in the form ln(ax+b) + ln(cx-d) + g*ln(1-x) where a, b, c, d and g are constants to be found. 7. It is given that e^(3y) * (1 - 4x) = 7 + 3x. Show that y = -1/3[ln(7+3x) - ln(1-4x)] 8. Find the value of x given that 3^(x+2) = 1/81 Read more

Steps to Solve

1. Solve for linear equations from the first equation

Given $\frac{2^x}{4^y} = 1$, we can simplify this using the property $4 = 2^2$: $\frac{2^x}{(2^2)^y} = 1$ $2^x = 2^{2y}$. Since the bases are equal, the exponents must be equal: $x = 2y$.

2. Solve the second exponential equation

Given $3^x(\sqrt{3})^y = 9$. Rewrite $\sqrt{3}$ as $3^{\frac{1}{2}}$ and $9$ as $3^2$: $3^x (3^{\frac{1}{2}})^y = 3^2$ $3^x \cdot 3^{\frac{y}{2}} = 3^2$ $3^{x + \frac{y}{2}} = 3^2$. Equating the exponents gives us: $x + \frac{y}{2} = 2$.

3. Solve for x and y using the system of equations

We have the system of equations: $x = 2y$ $x + \frac{y}{2} = 2$ Substitute $x = 2y$ into the second equation: $2y + \frac{y}{2} = 2$ $\frac{4y}{2} + \frac{y}{2} = 2$ $\frac{5y}{2} = 2$ $5y = 4$ $y = \frac{4}{5}$ Now, substitute $y = \frac{4}{5}$ back into $x = 2y$: $x = 2(\frac{4}{5})$ $x = \frac{8}{5}$

4. Solve the equation $\frac{e^{3x}-e^{-3x}}{2} = 4$

Multiply both sides by 2: $e^{3x} - e^{-3x} = 8$ Multiply through by $e^{3x}$ $e^{6x} - 1 = 8e^{3x}$ $e^{6x} - 8e^{3x} - 1 = 0$ Let $u = e^{3x}$ $u^2 - 8u - 1 = 0$

5. Use the quadratic formula to solve for $u$

$u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ $u = \frac{8 \pm \sqrt{(-8)^2 - 4(1)(-1)}}{2(1)}$ $u = \frac{8 \pm \sqrt{64 + 4}}{2}$ $u = \frac{8 \pm \sqrt{68}}{2}$ $u = \frac{8 \pm 2\sqrt{17}}{2}$ $u = 4 \pm \sqrt{17}$ Since $u = e^{3x}$, $u$ must be positive. So, $u = 4 + \sqrt{17}$

6. Solve for $x$

$e^{3x} = 4 + \sqrt{17}$ Take the natural logarithm of both sides: $3x = ln(4 + \sqrt{17})$ $x = \frac{ln(4 + \sqrt{17})}{3}$ $x \approx 0.764$ (to 3 s.f.)

7. Solve for $x$ using the equation $2e^{2x}-5e^x + 2 = 0$

Substitute $y = e^x$. $2y^2 - 5y + 2 = 0$ Factor the quadratic: $(2y - 1)(y - 2) = 0$ So, $2y - 1 = 0$ or $y - 2 = 0$ $y = \frac{1}{2}$ or $y = 2$ Since $y = e^x$: $e^x = \frac{1}{2}$ or $e^x = 2$ Taking the natural logarithm of both sides gives $x = ln(\frac{1}{2})$ or $x = ln(2)$ $x = -ln(2)$ or $x = ln(2)$

8. Solve for $x$ using the logarithmic equation $log_{10}(2x^3 + 1) - 3log_{10} x = 1$

Using logarithm properties, $3log_{10} x = log_{10} x^3$, so we rewrite the equation as: $log_{10}(2x^3 + 1) - log_{10} x^3 = 1$ $log_{10}(\frac{2x^3 + 1}{x^3}) = 1$ Convert to exponential form: $\frac{2x^3 + 1}{x^3} = 10^1$ $2x^3 + 1 = 10x^3$ $1 = 8x^3$ $x^3 = \frac{1}{8}$ $x = \sqrt[3]{\frac{1}{8}} = \frac{1}{2}$

9. Solve the system of simultaneous equations $2^x + 3^y = 5$ and $2^{x+2} - 3^{y+1} = 13$

Rewrite the second equation: $2^{x+2} - 3^{y+1} = 2^x \cdot 2^2 - 3^y \cdot 3^1 = 4 \cdot 2^x - 3 \cdot 3^y = 13$ Let $a = 2^x$ and $b = 3^y$. The system becomes: $a + b = 5$ $4a - 3b = 13$ From the first equation, $b = 5 - a$. Substitute into the second equation: $4a - 3(5 - a) = 13$ $4a - 15 + 3a = 13$ $7a = 28$ $a = 4$ So, $2^x = 4 = 2^2$, which means $x = 2$. Then, $b = 5 - a = 5 - 4 = 1$. So, $3^y = 1 = 3^0$, which means $y = 0$.

10. Express $ln(\frac{2x^2 - 3x - 2}{\sqrt{1-x}})$ in the form $ln(ax+b) + ln(cx-d) + gln(1-x)$

First, factor the quadratic: $2x^2 - 3x - 2 = (2x + 1)(x - 2)$ Then we have: $ln(\frac{(2x+1)(x-2)}{\sqrt{1-x}}) = ln((2x+1)(x-2)) - ln(\sqrt{1-x})$ $ = ln(2x+1) + ln(x-2) - \frac{1}{2}ln(1-x)$ So we have $a=2, b=1, c=1, d=2, g = -\frac{1}{2}$. $ln(2x+1) + ln(x-2) - \frac{1}{2}ln(1-x)$

11. Show that $y = -\frac{1}{3}[ln(7+3x) - ln(1-4x)]$ given $e^{3y}(1 - 4x) = (7+3x)$

$e^{3y} = \frac{7+3x}{1-4x}$ Take the natural log of both sides: $3y = ln(\frac{7+3x}{1-4x}) = ln(7+3x) - ln(1-4x)$ $y = \frac{1}{3}[ln(7+3x) - ln(1-4x)]$ $y = -\frac{1}{3}[ln(1-4x)- ln(7+3x)]$ The provided answer has a sign error

12. Find the value of x given that $3^{x+2} = \frac{1}{81}$ Express 1/81 as a power of 3 $\frac{1}{81} = \frac{1}{3^4} = 3^{-4}$ $3^{x+2} = 3^{-4}$ Since the bases are equal, equate the exponents: $x + 2 = -4$ $x = -6$