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Solve for \(t\). a. \(10^{t}=4\) d. \(10^{-t}=5\) b. \(3\left(2^{t}\right)=21\) e. \(5^{t}=7^{t+1}\) c. \(1+5^{t}=3\) f. \(6 \cdot 2^{t}=3^{t-1}\)

Short Answer

Expert verified
a) t ≈ 0.602, d) t ≈ -0.699, b) t ≈ 2.807, e) t ≈ -2.515, c) t ≈ 0.431, f) t ≈ 2.709.

Step by step solution

01

Title - Solving for t in Exercise (a)

Given the equation \(10^{t}=4\), we take the logarithm of both sides: \(\log(10^{t})=\log(4)\). By using the property of logarithms \(\log(b^{c})=c\log(b)\), we get: \(t\log(10)=\log(4)\). Since \(\log(10)=1\), it simplifies to \(t=\log(4)\). Therefore, \(t\approx0.602\).
02

Title - Solving for t in Exercise (d)

Given the equation \(10^{-t}=5\), we take the logarithm of both sides: \(\log(10^{-t})=\log(5)\). Using the property of logarithms \(\log(b^{-c})=-c\log(b)\), we get: \(-t\log(10)=\log(5)\). Since \(\log(10)=1\), it simplifies to \(-t=\log(5)\), so \(t=-\log(5)\). Therefore, \(t\approx-0.699\).
03

Title - Solving for t in Exercise (b)

Given the equation \(3\left(2^{t}\right)=21\), we divide both sides by 3: \(2^{t}=7\). By taking the logarithm of both sides, \(\log(2^{t})=\log(7)\). Using the property of logarithms, we get: \(t\log(2)=\log(7)\). So, \(t=\frac{\log(7)}{\log(2)}\). Therefore, \(t\approx 2.807\).
04

Title - Solving for t in Exercise (e)

Given the equation \(5^{t}=7^{t+1}\), we take the logarithm of both sides: \(\log(5^{t})=\log(7^{t+1})\). By using the property of logarithms, we get: \(t\log(5)=(t+1)\log(7)\). Expanding the right side, we have: \(t\log(5)=t\log(7)+\log(7)\). Solving for t, we get: \(t\log(5)-t\log(7)=\log(7)\). \(t(\log(5)-\log(7))=\log(7)\). Thus, \(t=\frac{\log(7)}{\log(5)-\log(7)}\). Therefore, \(t\approx -2.515\).
05

Title - Solving for t in Exercise (c)

Given the equation \(1+5^{t}=3\), we isolate the exponential term: \(5^{t}=2\). By taking the logarithm of both sides, \(\log(5^{t})=\log(2)\). Using the property of logarithms, we get: \(t\log(5)=\log(2)\). So, \(t=\frac{\log(2)}{\log(5)}\). Therefore, \(t \approx 0.431\).
06

Title - Solving for t in Exercise (f)

Given the equation \(6 \cdot 2^{t}=3^{t-1}\), we first simplify it. By isolating the exponential terms, we get: \(2^{t}=\frac{3^{t-1}}{6}\). Taking the logarithm of both sides, \(\log(2^{t})=\log\left(\frac{3^{t-1}}{6}\right)\). Using the property of logarithms, we get: \(t\log(2)=\log(3^{t-1})-\log(6)\). Expanding the right side, we have: \(t\log(2)=(t-1)\log(3)-\log(6)\). Distributing, we get: \(t\log(2)=t\log(3)-\log(3)-\log(6)\). Solving for t, \(t\log(2)-t\log(3)=-\log(3)-\log(6)\). \(t(\log(2)-\log(3))=-\log(3)-\log(6)\). Thus, \(t=\frac{-\log(3)-\log(6)}{\log(2)-\log(3)}\). Therefore, \(t\approx 2.709\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Logarithms
Logarithms are essential tools in algebra that help us solve exponential equations. They allow us to transform multiplicative relationships into additive ones, making equations easier to handle. The logarithm of a number is the power to which a base, usually 10 (common logarithm) or e (natural logarithm), must be raised to get that number. For instance, \(\rmlog(1000) = 3\) because \(\rm10^3 = 1000\). Logarithms are particularly useful for solving equations where the variable is an exponent.
Logarithmic Properties
Understanding the properties of logarithms can simplify many algebraic problems. Some key properties include:
  • Product Property: \(\rmlog_b(xy) = \rmlog_b(x) + \rmlog_b(y)\)
  • Quotient Property: \(\rmlog_b(x/y) = \rmlog_b(x) - \rmlog_b(y)\)
  • Power Property: \(\rmlog_b(x^c) = c\rmlog_b(x)\)
  • Change of Base Formula: \(\rmlog_b(x) = \rmlog_k(x) / \rmlog_k(b)\)

These properties allow for the manipulation of logarithmic expressions and are frequently utilized in solving exponential equations.
For example, when we encounter \(\rmlog(10^t) = t\rmlog(10)\), we use the power property to simplify it to just \(\rmt = \rmlog(4)\) when \(\rm10^t = 4\).
Exponential Functions
Exponential functions are mathematical expressions in which a constant base is raised to a variable exponent. They are in the form \(\rmf(x) = a^x\), where 'a' is a positive real number. These functions grow rapidly and are used to model many real-world phenomena, such as compound interest and population growth. When solving equations involving exponential functions, logarithms become particularly helpful. For instance, to solve \(\rm3(2^t) = 21\), we isolate the exponential term, divide by 3 to obtain \(\rm2^t = 7\), and then apply logarithms to both sides to find the value of 't'.

Exponential functions have specific characteristics:
  • They are always positive.
  • Their growth rate increases over time (when the base > 1).
  • They have a horizontal asymptote at y=0.

Understanding these properties aids in visualizing and solving equations.
Mathematical Problem-Solving
Effective problem-solving in mathematics involves a series of steps to break down complex problems into manageable parts. Here are some essential steps and strategies:
  • Identify and understand the problem.
  • Gather and write down known information.
  • Determine the best strategy to solve the problem (e.g., isolating the variable, using logarithms).
  • Execute the strategy step by step.
  • Always check your work to ensure accuracy.

Applying these steps to our examples, we find the value of 't' by isolating the exponential term and using logarithms to simplify the equation. For instance, with \(\rm5^t = 7^{t+1}\), we take the logarithm of both sides and apply logarithmic properties, which eventually leads us to the solution. This systematic approach not only ensures correct answers but also builds a stronger understanding of mathematical concepts.

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Most popular questions from this chapter

a. From the data in the following table, create a linear equation of \(Y\) in terms of \(x\). $$ \begin{array}{cc} \hline x & \log y \text { (or } Y \text { ) } \\ \hline 0 & 5.00000 \\ 1 & 5.60206 \\ 2 & 6.20412 \\ 3 & 6.80618 \\ 4 & 7.40824 \\ \hline \end{array} $$ b. Find the equivalent exponential function of \(y\) in terms of \(x\).

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Solve for \(x\). a. \(\log x=3\) d. \(\log (x+1)-\log x=1\) b. \(\log (x+1)=3\) e. \(\log x-\log (x+1)=1\) c. \(3 \log x=5\)

The half-life of bismuth-214 is about 20 minutes. a. Construct a function to model the decay of bismuth- 214 over time. Be sure to specify your variables and their units. b. For any given sample of bismuth- 214 , how much is left after I hour? c. How long will it take to reduce the sample to \(25 \%\) of its original size? d. How long will it take to reduce the sample to \(10 \%\) of its original size?

a. Phosphorus- 32 is used to mark cells in biological experiments. If phosphorus-32 has a continuous daily decay rate of 0.0485 or \(4.85 \%,\) what is its half-life? (Hint: Rewrite the function as \(y=\mathrm{Ca}^{x}\) and set \(y=0.5 \mathrm{C} .\) ) b. Phosphorus-32 can be quite dangerous to work with if the experimenter fails to use the proper shields, since its highenergy radiation extends out to \(610 \mathrm{~cm}\) or about 20 feet. Because disposal of radioactive wastes is increasingly difficult and expensive, laboratories often store the waste until it is within acceptable radioactive levels for disposal with non-radioactive trash. For instance, the rule of thumb for the laboratories of a large East Coast university and medical center is that any waste containing radioactive material with a half-life under 65 days must be stored for 10 half-lives before disposal with the non-radioactive trash. i. For how many days would phosphorus- 32 have to be stored? ii. What percentage of the original phosphorus-32 would be left at that time?

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