17Calculus Precalculus - Solve Exponentials

17Calculus

Solving Equations Involving Exponentials

When we have a variable inside a logarithm, we need to move it out of the logarithm to determine it's value. To do that, we use the laws listed on the laws of logarithms page. Here is how it works.

When we have an exponential, say $$e^x$$, to get the variable out of the exponent we take the natural log to get $$\ln (e^x)$$. Then we use the law $$\ln(x^y) = y\ln x$$. This gives us $$\ln (e^x) = x\ln e$$. But $$\ln e = 1$$, so we end up with $$\ln (e^x) = x\ln e = x$$. Let's go through an example to see this in action.

Example

Solve $$2^x = 30$$.

 Solution $$2^x = 30$$ Take the natural log of both sides. $$\ln(2^x) = \ln(30)$$ Use the law $$\ln(x^y) = y\ln x$$ to simplify the left side. $$x\ln(2) = \ln(30)$$ Solve for $$x$$. $$x = \ln(30)/\ln(2)$$

Practice

Unless otherwise instructed, solve these problems using the natural logarithm giving your answers in exact terms.

$$3^x = 5$$

Problem Statement

Solve $$3^x = 5$$

Solution

3035 video solution

Log in to rate this practice problem and to see it's current rating.

$$5^{2x+3} = 8$$

Problem Statement

Solve $$5^{2x+3} = 8$$

Solution

3036 video solution

Log in to rate this practice problem and to see it's current rating.

$$3^{x+2} = 4^{2-x}$$

Problem Statement

Solve $$3^{x+2} = 4^{2-x}$$

Solution

3037 video solution

Log in to rate this practice problem and to see it's current rating.

$$8^x=15$$

Problem Statement

Solve $$8^x=15$$ using the natural logarithm giving your answer in exact terms.

Solution

PatrickJMT - 1698 video solution

video by PatrickJMT

Log in to rate this practice problem and to see it's current rating.

$$7^x-1=4$$

Problem Statement

Solve $$7^x-1=4$$ using the natural logarithm giving your answer in exact terms.

Solution

MIP4U - 1701 video solution

video by MIP4U

Log in to rate this practice problem and to see it's current rating.

$$3(2^x)-2=13$$

Problem Statement

Solve $$3(2^x)-2=13$$ using the natural logarithm giving your answer in exact terms.

Solution

MIP4U - 1702 video solution

video by MIP4U

Log in to rate this practice problem and to see it's current rating.

$$(2/3)^x=5^{3-x}$$

Problem Statement

Solve $$(2/3)^x=5^{3-x}$$ using the natural logarithm giving your answer in exact terms.

Solution

MIP4U - 1703 video solution

video by MIP4U

Log in to rate this practice problem and to see it's current rating.

$$5^{x-3}=3^{2x+1}$$

Problem Statement

Solve $$5^{x-3}=3^{2x+1}$$ using the natural logarithm giving your answer in exact terms.

Solution

MIP4U - 1704 video solution

video by MIP4U

Log in to rate this practice problem and to see it's current rating.

$$1111=5(2^t)$$

Problem Statement

Solve $$1111=5(2^t)$$ using the natural logarithm giving your answer in exact terms.

Solution

Khan Academy - 1700 video solution

Log in to rate this practice problem and to see it's current rating.

$$\displaystyle{\left( \frac{4}{5} \right)^x = 6^{1-x}}$$

Problem Statement

Solve $$\displaystyle{\left( \frac{4}{5} \right)^x = 6^{1-x}}$$ using the natural logarithm giving your answer in exact terms.

Solution

PatrickJMT - 1699 video solution

video by PatrickJMT

Log in to rate this practice problem and to see it's current rating.

The exponential function $$V(x) = 25 (4/5)^x$$ is used to model the value of a car over time. Use the properties of logarithms and exponentials to rewrite the model in the form $$V(t) = 25e^{kt}$$.

Problem Statement

The exponential function $$V(x) = 25 (4/5)^x$$ is used to model the value of a car over time. Use the properties of logarithms and exponentials to rewrite the model in the form $$V(t) = 25e^{kt}$$.

$$V(t) = 25e^{t\ln(4/5)}$$

Problem Statement

The exponential function $$V(x) = 25 (4/5)^x$$ is used to model the value of a car over time. Use the properties of logarithms and exponentials to rewrite the model in the form $$V(t) = 25e^{kt}$$.

Solution

Notice that the two equations are the same except for the exponential terms. So we need to determine $$k$$ from $$(4/5)^x = e^{kt}$$. Although not stated in the problem, since the function names are both V, we can assume that $$x=t$$.

 $$(4/5)^x = e^{kt}$$ $$\ln[(4/5)^x] = \ln[e^{kt}]$$ $$x\ln[(4/5)] = kt\ln[e]$$ $$x\ln[(4/5)] = kt$$ $$x\ln[(4/5)] = kt$$ $$(x/t)\ln[(4/5)] = k$$ Since $$x=t$$, $$x/t = 1$$ $$\ln[(4/5)] = k$$

$$V(t) = 25e^{t\ln(4/5)}$$

Log in to rate this practice problem and to see it's current rating.

When using the material on this site, check with your instructor to see what they require. Their requirements come first, so make sure your notation and work follow their specifications.

DISCLAIMER - 17Calculus owners and contributors are not responsible for how the material, videos, practice problems, exams, links or anything on this site are used or how they affect the grades or projects of any individual or organization. We have worked, to the best of our ability, to ensure accurate and correct information on each page and solutions to practice problems and exams. However, we do not guarantee 100% accuracy. It is each individual's responsibility to verify correctness and to determine what different instructors and organizations expect. How each person chooses to use the material on this site is up to that person as well as the responsibility for how it impacts grades, projects and understanding of calculus, math or any other subject. In short, use this site wisely by questioning and verifying everything. If you see something that is incorrect, contact us right away so that we can correct it.