How do i solve the following equations giving the answers either exactly or to 3 significant figures?

a)3x + 5 = 2x – 1

b)(x^2 + 9) / 2 = 10

c)x^2 = 4 (x – 1)

d)2^x = 5

e)3^x+1 = 7

What do you not understand about thse euqations. They seem pretty simple. Please be dtailed.

To solve these equations, we need to isolate the variable (x) on one side of the equation. Let's go through each equation step by step:

a) 3x + 5 = 2x - 1

In this equation, our goal is to get all the terms with x on one side and the constant terms on the other side. Let's start solving it:

Step 1: Subtract 2x from both sides to eliminate the 2x term on the right side:
3x - 2x + 5 = 2x - 2x - 1
x + 5 = -1

Step 2: Subtract 5 from both sides to get x by itself:
x + 5 - 5 = -1 - 5
x = -6

So the solution to this equation is x = -6.

b) (x^2 + 9) / 2 = 10

To solve this equation, we first multiply both sides by 2 to get rid of the fraction:

2 * [(x^2 + 9) / 2] = 2 * 10
x^2 + 9 = 20

Next, subtract 9 from both sides:

x^2 + 9 - 9 = 20 - 9
x^2 = 11

To get the value of x, we can take the square root of both sides. However, note that we should consider both positive and negative square roots since x can have two possible values:

x = ±√11

The solution to this equation is x = ±√11, which means x can be approximately 3.317 or -3.317, rounded to 3 significant figures.

c) x^2 = 4(x - 1)

To solve this quadratic equation, let's distribute the 4 on the right side:

x^2 = 4x - 4

Next, move all terms to one side of the equation:

x^2 - 4x + 4 = 0

This is a quadratic equation in the form ax^2 + bx + c = 0, where a = 1, b = -4, and c = 4.

We can solve this quadratic equation by factoring, completing the square, or using the quadratic formula. In this case, we can factor it easily:

(x - 2)(x - 2) = 0
(x - 2)^2 = 0

Since the square of any real number is 0, we have:

x - 2 = 0

Adding 2 to both sides:

x = 2

So the solution to this equation is x = 2.

d) 2^x = 5

To solve this exponential equation, we need to take the logarithm of both sides. Let's use the natural logarithm (ln) in this case:

ln(2^x) = ln(5)

Using the property of logarithms, we can bring down the exponent:

x * ln(2) = ln(5)

Now, we divide both sides by ln(2) to isolate x:

x = ln(5) / ln(2)

Using a calculator, we can evaluate the right side to get the decimal approximation. In this case, x ≈ 2.32192, rounded to 3 significant figures.

e) 3^(x+1) = 7

Similar to the previous equation, we need to take the logarithm of both sides to solve for x:

ln(3^(x + 1)) = ln(7)

Again, using the property of logarithms, we can bring down the exponent:

(x + 1) * ln(3) = ln(7)

Now, to isolate x, let's divide both sides by ln(3):

x + 1 = ln(7) / ln(3)

Subtract 1 from both sides:

x = ln(7) / ln(3) - 1

Using a calculator, we can evaluate the right side to find the decimal approximation. In this case, x ≈ 1.13974, rounded to 3 significant figures.

So, those are the solutions to the given equations, either exactly or rounded to 3 significant figures.