3.7 Solving other Exponential and Logarithmic Equations
3.7.1 Classwork
Solve.
- [latex]\log_7(x-2)=1[/latex]
- [latex]\log_2(2x+6)=1[/latex]
- [latex]\log_6(x-2)=3[/latex]
- [latex]2\log_7(x-2)=1[/latex]
- [latex]\log_2(2x+6)+2=1[/latex]
- [latex]\log_6(x-2)-2=0[/latex]
Answer the following.
- How do you solve [latex]y=e^x[/latex] for x?
- How do you solve [latex]y=\log(x)[/latex] for x?
- How do you solve [latex]y=\log_3(x)[/latex] for x?
The Power Property for Logarithms: [latex]\log_b(x^p)=p\log_b(x)[/latex], where [latex]b>0[/latex]
Use the Power Property for Logarithms.
- [latex]\ln(x^3)[/latex]
- [latex]\log\left[{(x-2)}^5\right][/latex]
- [latex]\log_5\left(\sqrt x\right)[/latex]
- [latex]In\;\left(\sqrt[5]{x^2}\right)[/latex]
- [latex]\ln\left(b^c\right)[/latex]
Solve. Round approximate answers to three decimal places.
- [latex]250=500{(1-0.07)}^t[/latex]
- [latex]100=500\left(1-\frac{0.07}{12}\right)^{12t}[/latex]
- [latex]1,000=500{(1+0.01)}^t[/latex]
- [latex]\log_2(4x-1)=-3[/latex]
- [latex]\log_3(2x+1)=4[/latex]
- [latex]5\log_4(5-2x)=-10[/latex]
Solve.
If [latex]a^u=a^v[/latex], then [latex]u=v[/latex]. Convert each side of the equation to the same base, then solve.
- [latex]2^{x-4}=2^{3x+7}[/latex]
- [latex]3^{2x}=27^{x-1}[/latex]
- [latex]2^{2x-1}=32[/latex]
- [latex]5^x=625[/latex]
- [latex]4^{2x-1}=64[/latex]
- [latex]32^x=8[/latex]
- [latex]4^x=\frac1{16}[/latex]
- [latex]8=\frac1{4^x}[/latex]
3.7.2 Homework
Solve for the exact answer.
- [latex]10^x=34,926[/latex]
- [latex]e^x=64[/latex]
- [latex]3^7=x[/latex]
- [latex]4^x=1024[/latex]
- [latex]3\ln(5x)=9[/latex]
- [latex]e^{4x-5}-7=11,243[/latex]
- [latex]e^{1-8x}=7957[/latex]
- [latex]4e^{7x}=10,273[/latex]
- [latex]9^x=27[/latex]
- [latex]5^{2-x}=\frac1{125}[/latex]
- [latex]3^{1-x}=\frac1{27}[/latex]
- [latex]250=50e^{0.06\cdot t}[/latex]
- [latex]25=50e^{k7}[/latex]
- [latex]25=50e^{k10000}[/latex]
- [latex]1000=500{(1+.01)}^t[/latex]
- [latex]\log(4x)=2[/latex]
- [latex]5{(2)}^{3x}=20[/latex]
- [latex]4^{3x}=6[/latex]
- [latex]\log(2x+5)=2[/latex]
- [latex]\log(5x)=3[/latex]
Solve. Round approximate answers to two decimal places.
- If your money grew to [latex]\$12,000[/latex] invested at simple interest on an investment of [latex]\$10,000[/latex] in five years, what is the rate of interest?
- How long would it take your brand new car to depreciate [latex]40\%?[/latex] Assume that it depreciates at [latex]15\%[/latex] per year (compounded continuously).
(If you lost [latex]40\%[/latex], what would A=?) - The population of a town increases by [latex]P(t)=2500e^{0.0293t}[/latex] where t is time in years since 1990. Find the population in each of the following years.
- [latex]2000[/latex]
- [latex]2010[/latex]
- [latex]2018[/latex]
- The population of a city in 2000 was 240,360, and the exponential growth rate was [latex]1.2\%[/latex]. Write the exponential growth function. When will the population be 275,000?
- Bacteria growth is modeled by this function where N is the number of bacteria in thousands, and t is the time in minutes: [latex]N=e^{kt}[/latex] If the number of bacteria is 570,000 when time is 4 minutes, find the value for k, the rate of growth.
- Radioactive decay is modeled by the function where R is the ratio of remaining material, k is the rate of decay, and t is time in years: [latex]R=e^{kt}[/latex]
- Find k at 800 years when R=0.80 ([latex]80\%[/latex] of its radioactive energy).
- Find the half-life of this element. (Half-life means the material still has [latex]50\%[/latex] of its life left).
- At the end of five years, an initial investment of [latex]\$7,000[/latex] is now worth [latex]\$8,549.82[/latex]. Assuming that interest was compounded continuously, what was the rate?
- At the end of seven years, an initial investment of [latex]\$15,000[/latex] is now worth [latex]\$19,502.05[/latex]. Assuming that interest was compounded continuously, what was the rate?
- What would you have to invest in order to reach [latex]\$10,000[/latex] in 5 years at a rate of [latex]2.5\%[/latex], compounded continuously?
- If Free Yer Assets Bank (FYAB) will give you [latex]0.25\%[/latex] compounded quarterly and [latex]\text{I}.\text{M}.\text{A}.\text{Q}.\text{T}.\mathrm\pi[/latex] bank will give you [latex]0.23\%[/latex] interest compounded continuously, which is offering the better deal (Hint: Show how long it takes to double an investment).
- After exercising, your heart rate can be modeled by [latex]R\left(t\right)=151e^{-0.055t}[/latex] where [latex]0\leq t\leq15[/latex] and t is the number of minutes that have elapsed after you stop exercising.
- How many minutes does it take your heart rate to drop to 100 beats per minute?
- Find your heart rate 15 minutes after exercising.