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.
4^(x-1) = 3^2x

Question

Accepted Answer

Hey, everyone here we are asked to solve the following equation for X and round the answer to the nearest thousands. Here, our given equation is seven raised to the power of X minus eight is equal to four, raised to the power of five X. Here we have four answer choice options. Answer a negative 2.87, answer B negative 4.806, answer C negative 2.982 and answer D negative 3.122. So looking back at our given equation here, we can see that both X terms are in our exponents of some base. So we want to manipulate our given equation such that we get our X variables outside of the exponents. So to do this, we first just need to take the natural log of both sides of the equation. So doing this, we have the natural log of seven raised to the power of X minus A is equal to the natural log of four, raised to the power of five X. So from here to get our X terms 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 we can apply this property to both sides of our equation so far. And so now rewriting, we have the quantity of X minus eight times the natural log of seven is equal to five X times the natural log of four. So now our next step is to just factor in this natural log of seven on the left side of our equation. And so doing this, we now have X times the natural log of seven minus eight times. The natural log of seven is equal to five X times the natural log of four. So now we want to get all of our X terms on the same side of the equation. So we can just subtract five X times the natural log of four from both sides of our equation. And so now rewriting, we have X times the natural log of seven minus five X, a natural log of four minus eight times. The natural log of seven is equal to zero. Now, we want to manipulate this equation so far so that we get the X terms alone, which means first, we need to move this negative eight times a natural log of seven to the right side of the equation. And we do this by just adding eight times the natural log of seven to both sides of the equation. And so now rewriting, we have X times the natural log of seven minus five, X times the natural log of four is equal to eight times the natural log of seven. So now we see on the left side of the equation that both of our terms have the variable X, so we can just factor out X from both terms. And we get X times the quantity of the natural log of seven minus five. The log of four is equal to eight times the natural log of seven. So now to get our X term completely alone, we need to divide both sides by this set of parentheses where we have the natural log of seven minus five times the natural log of four. So again, we divide both sides by this expression and we end up with just X being equal to the expression where we have eight times the natural log of seven divided by the quantity of the natural log of seven minus five times the natural log of four. So now the last step here, since we have found the expression for X, we just want to approximate the solution by utilizing a calculator. So in our calculator, we just need to plug in this log of seven divided by the quantity of the natural log of seven minus five times the natural log of four. So then plugging in this expression into our calculators, we get the approximate solution of negative 3.122. So again, we have X is equal to negative 3.122. And this answer here leaves us with answer choice D where again, we have negative 3.122. 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. 4^(x-1) = 3^2x

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Key Concepts

Exponential Equations

Logarithms

Exact vs. Approximate Solutions