Solve each equation for the indicated variable. Use logarithms wi...

Question

Solve each equation for the indicated variable. Use logarithms with the appropriate bases. See Example 10. y = A + B(1 - e^(-Cx)), for x

Accepted Answer

Hey, everyone here we are asked to solve the given expression for X by making use of logarithms with proper bases. Here we are given the expression Y is equal to P plus seven Q times the quantity of one minus E raised to the power of, of negative nine R X. Here we have four answer choice options. Answer choice A X is equal to nine divided by R times the natural log of seven Q divided by the quantity of seven Q minus the quantity of Y minus P. Answer B X is equal to one divided by nine R times the natural log of Q divided by the quantity of seven times the quantity of seven Q minus the quantity of Y minus P. Answer C X is equal to one divided by nine R times a natural log of Q divided by the quantity of seven Q minus the quantity of Y minus P. And answer D X is equal to one divided by nine R times a natural log of seven Q divided by the quantity of seven Q minus the quantity of Y minus P. Now to begin solving for X in our given expression, we want to manipulate our given expression so that we get E alone since it contains a power including our X variable. So getting E alone, we can first just subtract P from both sides. So we have Y minus P is equal to seven Q times the quantity of one minus E raised to the power of negative nine R X. Now we can get rid of this coefficient of our quantity here. So we want to divide both sides by seven Q. And so now we have the quantity of Y minus P divided by seven Q is equal to one minus E raised to the power of negative nine R X. Now we can move over this one from the right side to the left side. So we have the quantity of Y minus P divided by seven Q minus one is equal to negative E raised to the power of negative nine R X. And so now we want to get E to be positive so we can simply multiply both sides by negative one. And so doing this, we now have E raised to the power of negative nine R X is equal to one minus the quantity of Y minus P divided by seven Q. And now we want to manipulate this current equation so that we can get X to be outside of this exponent. So first, we need to take the natural log of both sides. So in doing this, we have the natural log of E raised to the power of negative nine R X is equal to the natural log of one minus the quantity of Y minus P divided by seven Q. And now using the logarithmic identity, where the natural log of A raised to the power of X is equivalent to X times the natural log of A, we can rewrite the left side of our equation as negative nine R X times the natural log of E. And then we have the right side. So is equal to the natural log of one minus the quantity of Y minus P divided by seven Q. And so now the next simplification we can make is by recalling that the natural log of E is equivalent to one. So we really have negative nine R X is equal to the natural log of one minus the quantity of Y minus P divided by seven Q. And so now we can continue to get X alone by now dividing both sides by the coefficient of X which is negative nine R. So dividing both sides by negative nine R, we now have the equation X is equal to one divided by negative nine R times the natural log of one minus the quantity of Y minus P divided by seven Q. And now the next step is to simplify our equation by first combining both of the terms that are contained within the parentheses of the natural log So combining these terms, we have X is equal to one divided by negative nine R times the natural log of seven Q minus the quantity of Y minus P all divided by seven Q. And now we can do one last step to simplify this expression. And that includes by taking the negative term that is the coefficient of the natural log and bringing it as a power of the natural log by recalling the property from earlier where the natural log of A to the power of X is equal to X times the natural log of A. So utilizing this property, we now have X is equal to one divided by nine R times the natural log of seven Q minus the quantity of Y minus P all divided by seven Q and now raised to the power of negative one. So now simplifying this expression, we are left with X is equal to one divided by nine R times the natural log of seven Q divided by the quantity of seven Q minus the quantity of Y minus P. And so now we have fully simplified our expression for X. And so with this expression, we are left here with answer choice D where again we have X is equal to one divided by nine R MC natural log of seven Q divided by the quantity of seven Q minus the quantity of Y minus P. Thanks for watching. I hope you found this video helpful. And I'll see you again next time.

Struggling with College Algebra?

Solve each equation for the indicated variable. Use logarithms with the appropriate bases. See Example 10. y = A + B(1 - e^(-Cx)), for x

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