Exponential Problems

Question

Solve equations algebraically. Approximate the result to three decimals.

a. e^x = e^{x^2 - 2}

b. 7 - 2e^x = 5

Answer 1

Part a) e^x = e^{x^2 - 2}

To solve this, since both sides are exponentials with the same base e, we can take the natural log of both sides:

ln e^x = ln (e^{x^2 - 2})

This allows us to take the exponents on both sides and equate them like this:

x = x² - 2 We can move everything to one side by subtracting x on both sides

-x -x

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x² - x - 2 = 0 This is a quadratic (highest power is 2) and is in the standard form of quadratics ax²+bx+c

so it can be factored into:

(x - 2)(x+1) = 0

x = 2, x = -1

Part b) 7 - 2e^x = 5

To solve this, we can bring combine like terms by subtracting 7 on both sides since 5 and 7 are both constants:

7 - 2e^x = 5

-7 -7

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-2e^x = -2 Now, we can divide by -2 on both sides to isolate e^x on one side

-2e^x / -2 = -2 / -2

e^x = 1 We know that anything raised to the power of 0 gives us 1 so x = 0

x = 0

For part a, x = 2 and x = -1.

For part b, x = 0.

Answer 2

First question:

Notice the base of the exponents are both e.

Therefore, you can just set the powers of the two exponents equal to each other.

x = x² - 2

I hope you can solve this.

If not, please leave a comment.

Second question:

You must isolate the exponent, so let's start off by subtracting 7 to both sides.

-2e^x = -2

Now, divide both sides by 2 to isolate even further.

e^x = 1

Lastly, we want to 'bring down" the x from the exponent, so we will take the logarithm base e (ie ln) of both sides.

x = ln(1)

Since ln(1) = 0, we have

x = 0

*Please feel free to ask me any question for clarification.