Solve the exponential equation. 7 2 x 2 − 8 = 1 7^{2x^2-8}=1 7 2 x 2 − 8 = 1

Here we're asked to solve the exponential equation. And the equation we're given is 7 to the power of 2x squared minus 8 is equal to 1. Now it looks like it might be a little bit complicated, but let's go ahead and just start from the beginning, step 1, and isolate our exponential expression. Now our exponential expression is actually already isolated, so we can go ahead and move on to step number 2 and take the logs of both sides. Now looking at this, I do not have a base 10. I have a base of 7, so that tells me that I'm going to go ahead and take the natural log of both sides. So let's go ahead and do that. We get the natural log of 7 to the power of 2x squared minus 8, and that is equal to the natural log of 1. So step 2 is done. We can go ahead and move on to step 3 and use log rules to get x out of our exponent. So we can go ahead and pull all of this, this 2x squared minus 8, to the front of our log using that power rule. So this becomes 2x squared minus 8 times the natural log of 7 is equal to the natural log of 1. Now we can actually use one other log rule here. Whenever we take the log of any base of 1, we know that that's just going to be 0. So this natural log of 1 actually completely cancels out. It is just 0. So this 2x squared minus 8 times the natural log of 7 is actually all equal to just 0. Now we can go ahead and move on to step 4 and proceed with solving for x. Now the first thing I need to do here is get rid of this natural log of 7. Because I know that this is just a constant. I can go ahead and divide both sides by the natural log of 7. But since I'm dividing by with 0 on the top here, I know that this is just going to be 0 because 0 over anything is still 0. So this just becomes 2x squared minus 8 is equal to 0. And from here, we can go ahead and solve this. Now, we're actually dealing with a quadratic equation here. So let's go ahead and keep solving. Now, solving for x here, we're gonna add 8 to both sides. And then we're left with 2x squared is equal to 8. And then I can divide both sides by 2 as well, canceling out that 2, and I'm left with x squared is equal to 4. Now I actually don't even need a calculator to get to my final answer here because if I take the square root of both sides I know that the square root of 4 is just equal to plus or minus 2. So I actually have 2 possible answers that would work here, positive 2 or negative 2. So we're actually completely done here and our answer is that x could either be positive 2 or negative 2. It's fine to get 2 answers every once in a while it totally still works and it's still the correct answer. Thanks for watching and let me know if you have questions.

Solve the exponential equation. 7 2 x 2 − 8 = 1 7^{2x^2-8}=1 7 2 x 2 − 8 = 1

x = ± 2.83 x=\pm2.83 x = ± 2.83

6. Exponential and Logarithmic Functions

Solving Exponential and Logarithmic Equations

Precalculus

6. Exponential and Logarithmic Functions

Solving Exponential and Logarithmic Equations

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Multiple Choice

Solve the exponential equation.
$7^{2x^2-8}=1$

$x=0.51$

$x=\pm2$

$x=\pm2.83$

$x=2.23$

Verified step by step guidance

Start by analyzing the given exponential equation: $ 17^{2x^2 - 8} = 1 $. Recognize that any number raised to the power of 0 is 1, so set the exponent equal to 0: $ 2x^2 - 8 = 0 $.

Solve the equation $ 2x^2 - 8 = 0 $ by first adding 8 to both sides to isolate the term with $ x $: $ 2x^2 = 8 $.

Divide both sides of the equation by 2 to solve for $ x^2 $: $ x^2 = 4 $.

Take the square root of both sides to solve for $ x $. Remember that taking the square root introduces both positive and negative solutions: $ x = \pm \sqrt{4} $.

Simplify the square root to find the solutions: $ x = \pm 2 $. These are the values of $ x $ that satisfy the original equation.