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We can express the relationship between logarithmic form and its corresponding exponential form as follows:
Note that the base $\text{\hspace{0.17em}}b\text{\hspace{0.17em}}$ is always positive.
Because logarithm is a function, it is most correctly written as $\text{\hspace{0.17em}}{\mathrm{log}}_{b}(x),$ using parentheses to denote function evaluation, just as we would with $\text{\hspace{0.17em}}f(x).\text{\hspace{0.17em}}$ However, when the input is a single variable or number, it is common to see the parentheses dropped and the expression written without parentheses, as $\text{\hspace{0.17em}}{\mathrm{log}}_{b}x.\text{\hspace{0.17em}}$ Note that many calculators require parentheses around the $\text{\hspace{0.17em}}x.$
We can illustrate the notation of logarithms as follows:
Notice that, comparing the logarithm function and the exponential function, the input and the output are switched. This means $\text{\hspace{0.17em}}y={\mathrm{log}}_{b}\left(x\right)\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}y={b}^{x}\text{\hspace{0.17em}}$ are inverse functions.
A logarithm base $\text{\hspace{0.17em}}b\text{\hspace{0.17em}}$ of a positive number $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ satisfies the following definition.
For $\text{\hspace{0.17em}}x>0,b>0,b\ne 1,$
where,
Also, since the logarithmic and exponential functions switch the $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}y\text{\hspace{0.17em}}$ values, the domain and range of the exponential function are interchanged for the logarithmic function. Therefore,
Can we take the logarithm of a negative number?
No. Because the base of an exponential function is always positive, no power of that base can ever be negative. We can never take the logarithm of a negative number. Also, we cannot take the logarithm of zero. Calculators may output a log of a negative number when in complex mode, but the log of a negative number is not a real number.
Given an equation in logarithmic form $\text{\hspace{0.17em}}{\mathrm{log}}_{b}\left(x\right)=y,$ convert it to exponential form.
Write the following logarithmic equations in exponential form.
First, identify the values of $\text{\hspace{0.17em}}b,y,\text{and}x.\text{\hspace{0.17em}}$ Then, write the equation in the form $\text{\hspace{0.17em}}{b}^{y}=x.$
Here, $\text{\hspace{0.17em}}b=6,y=\frac{1}{2},\text{and}x=\sqrt{6.}\text{\hspace{0.17em}}$ Therefore, the equation $\text{\hspace{0.17em}}{\mathrm{log}}_{6}\left(\sqrt{6}\right)=\frac{1}{2}\text{\hspace{0.17em}}$ is equivalent to $$\text{\hspace{0.17em}}{6}^{\frac{1}{2}}=\sqrt{6}.$$
Here, $\text{\hspace{0.17em}}b=3,y=2,\text{and}x=9.\text{\hspace{0.17em}}$ Therefore, the equation $\text{\hspace{0.17em}}{\mathrm{log}}_{3}\left(9\right)=2\text{\hspace{0.17em}}$ is equivalent to $\text{\hspace{0.17em}}{3}^{2}=9.\text{\hspace{0.17em}}$
Write the following logarithmic equations in exponential form.
To convert from exponents to logarithms, we follow the same steps in reverse. We identify the base $\text{\hspace{0.17em}}b,$ exponent $\text{\hspace{0.17em}}x,$ and output $\text{\hspace{0.17em}}y.\text{\hspace{0.17em}}$ Then we write $\text{\hspace{0.17em}}x={\mathrm{log}}_{b}\left(y\right).$
Write the following exponential equations in logarithmic form.
First, identify the values of $\text{\hspace{0.17em}}b,y,\text{and}x.\text{\hspace{0.17em}}$ Then, write the equation in the form $\text{\hspace{0.17em}}x={\mathrm{log}}_{b}\left(y\right).$
Here, $\text{\hspace{0.17em}}b=2,$ $\text{\hspace{0.17em}}x=3,$ and $\text{\hspace{0.17em}}y=8.\text{\hspace{0.17em}}$ Therefore, the equation $\text{\hspace{0.17em}}{2}^{3}=8\text{\hspace{0.17em}}$ is equivalent to $\text{\hspace{0.17em}}{\mathrm{log}}_{2}(8)=3.$
Here, $\text{\hspace{0.17em}}b=5,$ $\text{\hspace{0.17em}}x=2,$ and $\text{\hspace{0.17em}}y=25.\text{\hspace{0.17em}}$ Therefore, the equation $\text{\hspace{0.17em}}{5}^{2}=25\text{\hspace{0.17em}}$ is equivalent to $\text{\hspace{0.17em}}{\mathrm{log}}_{5}(25)=2.$
Here, $\text{\hspace{0.17em}}b=10,$ $\text{\hspace{0.17em}}x=-4,$ and $\text{\hspace{0.17em}}y=\frac{1}{\mathrm{10,000}}.\text{\hspace{0.17em}}$ Therefore, the equation $\text{\hspace{0.17em}}{10}^{-4}=\frac{1}{\mathrm{10,000}}\text{\hspace{0.17em}}$ is equivalent to $\text{\hspace{0.17em}}{\text{log}}_{10}(\frac{1}{\mathrm{10,000}})=-4.$
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