/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 27 Solve the exponential equation a... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Solve the exponential equation algebraically. Approximate the result to three decimal places. $$8\left(10^{3 x}\right)=12$$

Short Answer

Expert verified
The approximation of x to three decimal places using a calculator to evaluate the logarithm expression results in \(x = 0.079\).

Step by step solution

01

Rearrange the equation

First, divide both sides of the equation by 8 to remove the multiplication and isolate the exponential part on one side of the equation: \[10^{3x} = \frac{12}{8}\]This simplifies to: \[10^{3x} = 1.5\]
02

Apply Logarithm

To remove the exponent, apply logarithms to both sides of the equation. The base of the logarithm will be 10 because the base of the exponent is 10. \[log_{10}\left(10^{3x}\right) = log_{10}(1.5)\] Since the base of the logarithm and the base of the exponent are same, the equation simplifies to \(3x = log_{10}(1.5)\).
03

Isolate the variable

Finally, to isolate x, divide each side of the equation by 3:\[x = \frac{{log_{10}(1.5)}}{3}\]

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Graphical Analysis Use a graphing utility to graph \(f\) and \(g\) in the same viewing window and determine which is increasing at the greater rate as \(x\) approaches + \(\infty\). What can you conclude about the rate of growth of the natural logarithmic function? (a) \(f(x)=\ln x, \quad g(x)=\sqrt{x}\) (b) \(f(x)=\ln x, \quad g(x)=\sqrt[4]{x}\)

Find the domain, \(x\) -intercept, and vertical asymptote of the logarithmic function and sketch its graph. $$g(x)=\ln (-x)$$

Home Mortgage \(A \$ 120,000\) home mortgage for 30 years at \(7 \frac{1}{2} \%\) has a monthly payment of \(\$ 839.06\) Part of the monthly payment covers the interest charge on the unpaid balance, and the remainder of the payment reduces the principal. The amount paid toward the interest is $$u=M-\left(M-\frac{P r}{12}\right)\left(1+\frac{r}{12}\right)^{12 t}$$ and the amount paid toward the reduction of the principal is $$v=\left(M-\frac{P r}{12}\right)\left(1+\frac{r}{12}\right)^{12 t}$$ In these formulas, \(P\) is the size of the mortgage, \(r\) is the interest rate, \(M\) is the monthly payment, and \(t\) is the time (in years). (a) Use a graphing utility to graph each function in the same viewing window. (The viewing window should show all 30 years of mortgage payments.) (b) In the early years of the mortgage, is the greater part of the monthly payment paid toward the interest or the principal? Approximate the time when the monthly payment is evenly divided between interest and principal reduction. (c) Repeat parts (a) and (b) for a repayment period of 20 years \((M=\$ 966.71) .\) What can you conclude?

Use the acidity model given by \(\mathbf{p H}=-\log \left[\mathbf{H}^{+}\right],\) where acidity \((\mathbf{p H})\) is a measure of the hydrogen ion concentration \(\left[\mathbf{H}^{+}\right]\) (measured in moles of hydrogen per liter) of a solution. The \(\mathrm{pH}\) of a solution decreases by one unit. By what factor does the hydrogen ion concentration increase?

Find the domain, \(x\) -intercept, and vertical asymptote of the logarithmic function and sketch its graph. $g(x)=\log _{6} x$$

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.