The idea of logarithms is not as complicated as it might seem. Logarithms are just another way to write exponents. It's all about notation. The rules that apply to logarithms can be understood if you keep in mind that you are working with exponents.
If you want a complete lecture on this topic, we recommend this video from one of our favorite instructors.
video by Prof Leonard 

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Getting Started
You should already be familiar with this expression \(\displaystyle{ 2^3 = 8 }\). This same expression written as a logarithm is \(\displaystyle{ 3 = \log_2 8 }\) and is read 'three is the logarithm (think:exponent) base two of eight' or 'three is log (think:power) eight base two'.
You could also say, if I have a base two and I want to get eight, what should the exponent (or power) of the base two be to get eight? The answer is three.
Let's pause for a minute and watch an interesting video talking about a unique way of looking at exponentials and logarithms.
video by 3Blue1Brown 

It would be nice if all teachers used the triangle idea when thinking about exponentials and logarithms but things won't change overnight. So we need to study and be able to use the traditional way of looking at logarithms.
In calculus, you will work mostly with logarithms with base \(e\). These are special logarithms called natural logarithms. The notation is a bit different. Instead of \( \log_e x \), you will need to write \(\ln(x) \) or \( \ln ~x \). It is considered incorrect notation to write \( \log_e x \).
What Are Logarithms?
Before we go on, here is a great video for you that explains what logarithms are and how they work. It is well worth your time to watch it.
video by Dr Chris Tisdell 

Some Logarithm Laws
Here are some laws you need to use when combining logarithms. We use the natural log here because it is the most common logarithm that you will use in calculus but the same rules apply regardless of the type logarithm.
1. \( \ln(xy) = \ln(x) + \ln(y) \)
2. \( \ln(x/y) = \ln(x)  \ln(y) \)
3. \( \ln(x^y) = y \ln(x) \)
4. \( e^{\ln(x)} = x \)
5. \( \ln(e) = 1 \)
Let's compare the first couple of laws to exponents.
1. \( e^x e^y = e^{x+y} \)
2. \(\displaystyle{ \frac{e^x}{e^y} = e^{xy} }\)
Look at the exponents in these two equations and compare them with the corresponding logarithm law. Do you see the similarities? Spend some time comparing them. Write them next to each other on a piece of paper. Thinking about them and turning them over in your mind repeatedly will help you really understand them and know how to use them.
Here is a good video that proves some logarithm properties. It will help you to understand them and how to use them.
video by PatrickJMT 

Okay, let's try some practice problems.
Unless otherwise instructed, simplify these logarithms using the laws above.
What is the domain of the logarithm function \( y = \log_2(5x) + 3 \)?
Problem Statement 

What is the domain of the logarithm function \( y = \log_2(5x) + 3 \)?
Solution 

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Condense \( 3\log x 5 \log y + (2/3)\log z \) into a single logarithm.
Problem Statement 

Condense \( 3\log x 5 \log y + (2/3)\log z \) into a single logarithm.
Hint 

Hint: Remember that when \(\log\) is written without a subscript, the base is assumed to be base 10.
Problem Statement 

Condense \( 3\log x 5 \log y + (2/3)\log z \) into a single logarithm.
Hint 

Hint: Remember that when \(\log\) is written without a subscript, the base is assumed to be base 10.
Solution 

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Expand \(\displaystyle{ \log \left[ \frac{\sqrt[3]{ab^2}}{c^4} \right]^5 }\) into a sum and/or difference of logarithms.
Problem Statement 

Expand \(\displaystyle{ \log \left[ \frac{\sqrt[3]{ab^2}}{c^4} \right]^5 }\) into a sum and/or difference of logarithms.
Solution 

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Simplify \(\log_2 80\log_2 5\)
Problem Statement 

Simplify \(\log_2 80\log_2 5\)
Solution 

video by Krista King Math 

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Simplify \( \log_3 18 + \log_3 4.5 \) using the rules of logarithms.
Problem Statement 

Simplify \( \log_3 18 + \log_3 4.5 \) using the rules of logarithms.
Solution 

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Simplify \( \ln e^{10}  \ln e^4 \) using the rules of logarithms.
Problem Statement 

Simplify \( \ln e^{10}  \ln e^4 \) using the rules of logarithms.
Solution 

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Simplify \( \log_4(256/64) \)
Problem Statement 

Simplify \( \log_4(256/64) \)
Solution 

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Simplify \( \log_2(128/8) \)
Problem Statement 

Simplify \( \log_2(128/8) \)
Solution 

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Simplify \(\displaystyle{ \log_2 \left[ \frac{128 \cdot 64}{8 \cdot 16} \right]^5 }\)
Problem Statement 

Simplify \(\displaystyle{ \log_2 \left[ \frac{128 \cdot 64}{8 \cdot 16} \right]^5 }\)
Solution 

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Simplify \( \log_2(16 \cdot 32)  \log_3(9 \cdot 27) \)
Problem Statement 

Simplify \( \log_2(16 \cdot 32)  \log_3(9 \cdot 27) \)
Hint 

Be careful to notice that the bases of the two logarithms are different.
Problem Statement 

Simplify \( \log_2(16 \cdot 32)  \log_3(9 \cdot 27) \)
Hint 

Be careful to notice that the bases of the two logarithms are different.
Solution 

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Graphing Logarithms
Here is a short video discussing how to graph logarithm functions using an example. It is important to have an idea what a logarithm graph looks like. You will need to know this when working with continuity in calculus.
video by PatrickJMT 

Solving Equations Involving Logarithms
When we have a variable in an exponent, we need to move it out of the exponent to determine it's value. To do that, we use the laws listed above. See the practice problems below for examples, then try a few on your own.
Unless otherwise instructed, solve these problems using the natural logarithm giving your answers in exact terms.
Solve \( \log_2(x+1) + \log_2(5x+1) = 6 \)
Problem Statement 

Solve \( \log_2(x+1) + \log_2(5x+1) = 6 \)
Solution 

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Solve \( \ln x = 7 \)
Problem Statement 

Solve \( \ln x = 7 \)
Solution 

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Solve \( \log_3(10x+1)  \log_3(x+1) = 2 \)
Problem Statement 

Solve \( \log_3(10x+1)  \log_3(x+1) = 2 \)
Solution 

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Solve \( 3^x = 5 \)
Problem Statement 

Solve \( 3^x = 5 \)
Solution 

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Solve \( 5^{2x+3} = 8 \)
Problem Statement 

Solve \( 5^{2x+3} = 8 \)
Solution 

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Solve \( 3^{x+2} = 4^{2x} \)
Problem Statement 

Solve \( 3^{x+2} = 4^{2x} \)
Solution 

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Solve \( \log_2 16 = x \)
Problem Statement 

Solve \( \log_2 16 = x \)
Solution 

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Solve \( \log_x 81 = 4 \)
Problem Statement 

Solve \( \log_x 81 = 4 \)
Solution 

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Solve \( \log_{32} x = 4/5 \)
Problem Statement 

Solve \( \log_{32} x = 4/5 \)
Solution 

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Solve \( \log_3(5x+1) = 4 \)
Problem Statement 

Solve \( \log_3(5x+1) = 4 \)
Solution 

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Solve \( \log x = 24 \)
Problem Statement 

Solve \( \log x = 24 \)
Solution 

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Solve \( \log_7 (x^2+3x+9) = 2 \)
Problem Statement 

Solve \( \log_7 (x^2+3x+9) = 2 \)
Solution 

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Solve \( \ln(3x2) = 5 \)
Problem Statement 

Solve \( \ln(3x2) = 5 \)
Solution 

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Solve \( 4\ln(2x1) + 3 = 11 \)
Problem Statement 

Solve \( 4\ln(2x1) + 3 = 11 \)
Solution 

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Solve \( \log_3 (5x+2) = \log_3 (7x8) \)
Problem Statement 

Solve \( \log_3 (5x+2) = \log_3 (7x8) \)
Solution 

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Solve \( \log_2 (x^2+4x) = \log_2 (5) \)
Problem Statement 

Solve \( \log_2 (x^2+4x) = \log_2 (5) \)
Solution 

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Solve \( \log_2 x + \log_2 (x+4) = 5 \)
Problem Statement 

Solve \( \log_2 x + \log_2 (x+4) = 5 \)
Solution 

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Solve \( \log_3 (x+1) = 3  \log_3 (x=7) \)
Problem Statement 

Solve \( \log_3 (x+1) = 3  \log_3 (x=7) \)
Solution 

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Solve \( \log_4 (2x+6)  \log_4 (x1) = 1 \)
Problem Statement 

Solve \( \log_4 (2x+6)  \log_4 (x1) = 1 \)
Solution 

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Solve \( \log_2 (x+3) = 4 + \log_2 (x3) \)
Problem Statement 

Solve \( \log_2 (x+3) = 4 + \log_2 (x3) \)
Solution 

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Unless otherwise instructed, solve \(8^x=15\) using the natural logarithm giving your answer in exact terms.
Problem Statement 

Unless otherwise instructed, solve \(8^x=15\) using the natural logarithm giving your answer in exact terms.
Solution 

video by PatrickJMT 

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Unless otherwise instructed, solve \(7^x1=4\) using the natural logarithm giving your answer in exact terms.
Problem Statement 

Unless otherwise instructed, solve \(7^x1=4\) using the natural logarithm giving your answer in exact terms.
Solution 

video by MIP4U 

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Unless otherwise instructed, solve \(3(2^x)2=13\) using the natural logarithm giving your answer in exact terms.
Problem Statement 

Unless otherwise instructed, solve \(3(2^x)2=13\) using the natural logarithm giving your answer in exact terms.
Solution 

video by MIP4U 

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Unless otherwise instructed, solve \((2/3)^x=5^{3x}\) using the natural logarithm giving your answer in exact terms.
Problem Statement 

Unless otherwise instructed, solve \((2/3)^x=5^{3x}\) using the natural logarithm giving your answer in exact terms.
Solution 

video by MIP4U 

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Unless otherwise instructed, solve \(5^{x3}=3^{2x+1}\) using the natural logarithm giving your answer in exact terms.
Problem Statement 

Unless otherwise instructed, solve \(5^{x3}=3^{2x+1}\) using the natural logarithm giving your answer in exact terms.
Solution 

video by MIP4U 

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Unless otherwise instructed, solve \(1111=5(2^t)\) using the natural logarithm giving your answer in exact terms.
Problem Statement 

Unless otherwise instructed, solve \(1111=5(2^t)\) using the natural logarithm giving your answer in exact terms.
Solution 

video by Khan Academy 

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Unless otherwise instructed, solve \(\displaystyle{\left( \frac{4}{5} \right)^x = 6^{1x}}\) using the natural logarithm giving your answer in exact terms.
Problem Statement 

Unless otherwise instructed, solve \(\displaystyle{\left( \frac{4}{5} \right)^x = 6^{1x}}\) using the natural logarithm giving your answer in exact terms.
Solution 

video by PatrickJMT 

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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 \( V(t) = 25e^{kt} \).
Final Answer 

\(V(t) = 25e^{t\ln(4/5)}\)
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 \(x=t\), \(x/t = 1\) 
\( \ln[(4/5)] = k \) 
Final Answer 

\(V(t) = 25e^{t\ln(4/5)}\)
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Advanced
Solve \( \log x^{\log x} = 49 \)
Problem Statement 

Solve \( \log x^{\log x} = 49 \)
Solution 

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Solve \( \log x^2 = (\log x)^2 \)
Problem Statement 

Solve \( \log x^2 = (\log x)^2 \)
Solution 

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Solve \( \log (\log x) = 4 \)
Problem Statement 

Solve \( \log (\log x) = 4 \)
Solution 

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Next
Okay, after working those practice problems, you are ready to tackle some application problems involving exponentials and logarithms, starting with exponential growth and decay.
Here is a playlist of the videos on this page.
Really UNDERSTAND Precalculus
The Unit Circle
The Unit Circle [wikipedia]
Basic Trig Identities
Set 1  basic identities  

\(\displaystyle{ \tan(t) = \frac{\sin(t)}{\cos(t)} }\) 
\(\displaystyle{ \cot(t) = \frac{\cos(t)}{\sin(t)} }\) 
\(\displaystyle{ \sec(t) = \frac{1}{\cos(t)} }\) 
\(\displaystyle{ \csc(t) = \frac{1}{\sin(t)} }\) 
Set 2  squared identities  

\( \sin^2t + \cos^2t = 1\) 
\( 1 + \tan^2t = \sec^2t\) 
\( 1 + \cot^2t = \csc^2t\) 
Set 3  doubleangle formulas  

\( \sin(2t) = 2\sin(t)\cos(t)\) 
\(\displaystyle{ \cos(2t) = \cos^2(t)  \sin^2(t) }\) 
Set 4  halfangle formulas  

\(\displaystyle{ \sin^2(t) = \frac{1\cos(2t)}{2} }\) 
\(\displaystyle{ \cos^2(t) = \frac{1+\cos(2t)}{2} }\) 
Trig Derivatives
\(\displaystyle{ \frac{d[\sin(t)]}{dt} = \cos(t) }\) 
\(\displaystyle{ \frac{d[\cos(t)]}{dt} = \sin(t) }\)  
\(\displaystyle{ \frac{d[\tan(t)]}{dt} = \sec^2(t) }\) 
\(\displaystyle{ \frac{d[\cot(t)]}{dt} = \csc^2(t) }\)  
\(\displaystyle{ \frac{d[\sec(t)]}{dt} = \sec(t)\tan(t) }\) 
\(\displaystyle{ \frac{d[\csc(t)]}{dt} = \csc(t)\cot(t) }\) 
Inverse Trig Derivatives
\(\displaystyle{ \frac{d[\arcsin(t)]}{dt} = \frac{1}{\sqrt{1t^2}} }\) 
\(\displaystyle{ \frac{d[\arccos(t)]}{dt} = \frac{1}{\sqrt{1t^2}} }\)  
\(\displaystyle{ \frac{d[\arctan(t)]}{dt} = \frac{1}{1+t^2} }\) 
\(\displaystyle{ \frac{d[\arccot(t)]}{dt} = \frac{1}{1+t^2} }\)  
\(\displaystyle{ \frac{d[\arcsec(t)]}{dt} = \frac{1}{\abs{t}\sqrt{t^2 1}} }\) 
\(\displaystyle{ \frac{d[\arccsc(t)]}{dt} = \frac{1}{\abs{t}\sqrt{t^2 1}} }\) 
Trig Integrals
\(\int{\sin(x)~dx} = \cos(x)+C\) 
\(\int{\cos(x)~dx} = \sin(x)+C\)  
\(\int{\tan(x)~dx} = \ln\abs{\cos(x)}+C\) 
\(\int{\cot(x)~dx} = \ln\abs{\sin(x)}+C\)  
\(\int{\sec(x)~dx} = \) \( \ln\abs{\sec(x)+\tan(x)}+C\) 
\(\int{\csc(x)~dx} = \) \( \ln\abs{\csc(x)+\cot(x)}+C\) 
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