Solve the following exponential equations: a) 49^x+1 + 72x = 350 b) 16^x = 84x−3 c) ( rac127)^m imes (81)⁻¹ = 243

Rewrite the terms with a common base of 7.

$$(7^2)^{x+1} + 7^{2x} = 350$$ $$7^{2(x+1)} + 7^{2x} = 350$$ $$7^{2x+2} + 7^{2x} = 350$$

Apply the exponent rule $a^{m+n} = a^m \cdot a^n$.

$$7^{2x} \cdot 7^2 + 7^{2x} = 350$$ $$49 \cdot 7^{2x} + 7^{2x} = 350$$

Factor out $7^{2x}$.

$$7^{2x}(49 + 1) = 350$$ $$50 \cdot 7^{2x} = 350$$

Divide by 50.

$$7^{2x} = \frac{350}{50}$$ $$7^{2x} = 7$$

Since the bases are the same, equate the exponents.

$$2x = 1$$ $$x = \frac{1}{2}$$

The value of $x$ is \boxed{\textbf{\frac{1}{2}}}.

Rewrite the terms with a common base of 2.

$$(2^4)^x = (2^3)^{4x-3}$$ $$2^{4x} = 2^{3(4x-3)}$$ $$2^{4x} = 2^{12x-9}$$

Since the bases are the same, equate the exponents.

$$4x = 12x - 9$$

Subtract $4x$ from both sides.

$$0 = 8x - 9$$

Add 9 to both sides.

$$9 = 8x$$

Divide by 8.

$$x = \frac{9}{8}$$

The value of $x$ is \boxed{\textbf{\frac{9}{8}}}.

Rewrite the terms with a common base of 3.

$$\left(3^{-3}\right)^m \times (3^4)^{-1} = 3^5$$

Apply the exponent rule $(a^b)^c = a^{bc}$.

$$3^{-3m} \times 3^{-4} = 3^5$$

Apply the exponent rule $a^m \cdot a^n = a^{m+n}$.

$$3^{-3m-4} = 3^5$$

Since the bases are the same, equate the exponents.

$$-3m - 4 = 5$$

Add 4 to both sides.

$$-3m = 9$$

Divide by -3.

$$m = \frac{9}{-3}$$ $$m = -3$$

The value of $m$ is $\boxed{\textbf{-3}}$.

First, express the equation in terms of P.

$$3^{2y-1} + 2 \times 3^{y-1} = 1$$

Rewrite the terms using $3^{y}$ and $3^{-1}$.

$$3^{2y} \cdot 3^{-1} + 2 \times 3^y \cdot 3^{-1} = 1$$ $$(3^y)^2 \cdot \frac{1}{3} + 2 \cdot 3^y \cdot \frac{1}{3} = 1$$

Substitute $P = 3^y$.

$$P^2 \cdot \frac{1}{3} + 2 \cdot P \cdot \frac{1}{3} = 1$$ $$\frac{P^2}{3} + \frac{2P}{3} = 1$$

Multiply the entire equation by 3.

$$P^2 + 2P = 3$$

Rearrange into a quadratic equation.

$$P^2 + 2P - 3 = 0$$

The equation in terms of P is \boxed{\textbf{P^2 + 2P - 3 = 0}}.

Next, find the value of $y$.
Solve the quadratic equation for P.

$$P^2 + 2P - 3 = 0$$

Factor the quadratic equation.

$$(P+3)(P-1) = 0$$

This gives two possible values for P:

$$P = -3 \quad \text{or} \quad P = 1$$

Recall that $P = 3^y$.
Case 1: $3^y = -3$.
There is no real solution for $y$ because $3^y$ is always positive.

Case 2: $3^y = 1$.
Rewrite 1 as $3^0$.

$$3^y = 3^0$$

Equate the exponents.

$$y = 0$$

The value of $y$ is $\boxed{\textbf{0}}$.

Simplify the expression inside the square root first.

$$2^x \times 5^{2x} \div 2^x$$ $$= \frac{2^x \times 5^{2x}}{2^x}$$

Cancel out $2^x$.

$$= 5^{2x}$$

Now, take the square root of the simplified expression.

$$\sqrt{5^{2x}} = (5^{2x})^{\frac{1}{2}}$$

Apply the exponent rule $(a^b)^c = a^{bc}$.

$$= 5^{2x \times \frac{1}{2}}$$ $$= 5^x$$

The simplified expression is \boxed{\textbf{5^x}}.