Solve each equation. In Exercises 11–34, give irrational solution...

Question

Solve each equation. In Exercises 11–34, give irrational solutions as decimals correct to the nearest thousandth. In Exercises 35-40, give solutions in exact form. See Examples 1–4.
6^(x+1) = 4^(2x-1)

Accepted Answer

Hey, everyone. Here we are asked to solve the following equation for X and round the answer to the nearest thousands. Here, the given equation is 11 raised to the power of X plus two is equal to nine. Raise to the power of three X minus seven. Here we have four answer choice options. Answer a 5.337, answer B 6.92, answer C 4.811 and answer D 5.812. Looking at our given equation for X, we see that both 11 and nine are raised to the power of some expression that includes X. So we want to first manipulate this equation so that we get X to be outside of the exponents. And so that means we first need to just take the natural log of both sides of our equation. So doing this, we have the natural log of 11 raised to the power of X plus two is equal to the natural log of nine, raised to the power of three X minus seven. And so now to get our X values or X variables to be outside of the exponents, we need to recall the logarithmic property which states that the natural log of X raised to the power of A is equivalent to a times the natural log of X. So now apply this property to both sides of the equation that we have. So far, we can rewrite our equation as the quantity of X plus two times the natural log of 11 is equal to the quantity of three X minus seven times the natural log of nine. So now we just want to factor in the natural log of 11 of 11 on the left side of our equation as well as factor in the natural log of nine on the right side of our equation. And so rewriting, we have X times the natural log of 11 plus two times Z natural log of 11 is equal to three X times the natural log of nine minus seven times the natural log of nine. So now we want to get our X terms to be on the same side of the equation. So we can begin by just subtracting three X times the natural log of nine from both sides of the equation. And so now again, rewriting, we have X times the natural log of 11 minus natural log of nine plus two times. The natural log of 11 is equal to negative seven times the natural log of nine. So now working to get our X terms alone, we first want to move this positive two times the natural log of to the right side of our equation. So we just need to subtract two times the natural log of from both sides of our equation. And once again, rewriting, we now have X times the natural log of 11 minus three, X times the natural log of nine is equal to negative seven times the natural log of nine minus two times the natural log of 11. So now on the left side of our equation, we see both terms have an X variable. So we can just factor out X from both terms. And we get X times the quantity of the natural log of minus three times the natural log of nine. And this expression is still equal to negative seven times the natural log of nine minus two times the natural log of 11. So now the next step is to finish getting X completely alone. And so we just need to divide both sides of the equation by the expression in the sets of parentheses on the left side. So we divide both sides by the expression, the natural log of 11 fancy natural log of nine. And again, we do this, it's of our equation. And so now rewriting, we see we have X alone now, so we have X is equivalent to the expression where we have the quantity of negative seven times the natural log of nine minus two times the natural log of 11 all divided by the natural log of 11 minus three times the natural log of nine. And so now, the last step here is to plug in this expression that we found for X into our calculator to approximate our solution. So we need to plug in the quantity of negative seven times the natural log of nine minus two times the natural log of 11 all divided by the natural log of 11 minus three times the natural log of nine. So plugging in this expression into our calculator, we get the approximate solution for X as 4.811. So with this final value for X, we are left here with answer choice C where again our solution is 4.811. Thanks for watching. I hope you found this video helpful and I'll see you again next time.

Solve each equation. In Exercises 11–34, give irrational solutions as decimals correct to the nearest thousandth. In Exercises 35-40, give solutions in exact form. See Examples 1–4. 6^(x+1) = 4^(2x-1)

Key Concepts

Exponential Equations

Logarithms

Exact vs. Approximate Solutions

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