# Logarithmic Function Calculator

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Instructions:
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Use this step-by-step Logarithmic Function Calculator, to find the logarithmic function that passes through two given points in the plane XY. You need to provide the points \((t_1, y_1)\) and \((t_2, y_2)\), and this calculator will estimate the appropriate exponential function and will provide its graph.

## Logarithmic Function Calculator from Two Points

The main purpose of this calculator is to estimate the parameters \(A_0\) and \(k\) for the logarithmic function \(f(t)\) which is defined as:

\[f(t) = A_0 \ln(k t)\]The parameters need to be so the logarithmic function passes through the two given points \((t_1, y_1)\) and \((t_2, y_2)\).

### How do you estimate a logarithmic function from two points?

Algebraically speaking, you need to solve the following system of equations to find the parameters \(A_0\) and \(k\):

\[y_1 = A_0 \ln(k t_1)\] \[y_2 = A_0 \ln(k t_2)\]By solving this system for the unknowns \(A_0\) and \(k\), we can find unique solutions, as long as \(t_1 = \not t_2\).

Indeed, by subtracting both sides of the equations:

\[\displaystyle y_1 - y_2 = A_0 \left( \ln(k t_1) - \ln(k t_2) \right)\] \[\displaystyle \Rightarrow \, y_1 - y_2 = A_0 \ln \left(\displaystyle\frac{k t_1}{k t_2}\right) \] \[\displaystyle \Rightarrow \, y_1 - y_2 = A_0 \ln \left(\displaystyle\frac{t_1}{t_2}\right) \] \[ \Rightarrow \, A_0 = \displaystyle \frac{y_1 - y_2}{\ln(t_1) - \ln(t_2)} \]which solves the equations for \(A_0\). Now, in order to solve for \(k\) we use the first equation and apply exponential to both sides::

\[y_1 = A_0 \ln(k t_1)\] \[ \Rightarrow \, \displaystyle e^{\frac{y_1}{A_0}} = k t_1 \] \[ \Rightarrow \, k = \displaystyle \frac{e^{\frac{y_1}{A_0}}}{t_1} \]and there we have found \(k\), as function of \(A_0\) that is already determined and known.

### How do you calculate an exponential function?

If instead of a logarithmic function you are interested in exponential behavior, then you should probably use this exponential function calculator , which follows the same logic of estimating parameters to enforce the function passing through two given points.