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Solve $\log_{5-x}{(x^2-2x+65)} = 2$

In this problem, it is given that the logarithm of a quadratic expression $x$ square minus $2$ times $x$ plus $65$ to base $5$ minus $x$ is equal to $2$.

$\log_{5-x}{(x^2-2x+65)}$ $\,=\,$ $2$

Now, let’s learn how to solve this logarithmic equation to solve the value of $x$.

Release the equation from Logarithms

It is essential to release the equation from the logarithmic form for solving the given logarithmic equation. So, use the mathematical relation between the exponents and logarithms for converting the logarithmic equation into exponential equation.

$\implies$ $x^2-2x+65$ $\,=\,$ $(5-x)^2$

Expand the square of difference of terms

On the right hand side of the equation, there is an expression and it is the square of $5$ minus $x$. The $5$ minus $x$ whole square can be expanded as per the square of difference formula.

$\implies$ $x^2$ $-$ $2x$ $+$ $65$ $\,=\,$ $5^2$ $+$ $x^2$ $-$ $2 \times 5 \times x$

Solve the equation by simplification

There is a quadratic expression on both sides of the equation. Move the expression on one side to other side for solving the equation by simplification.

$\implies$ $x^2$ $-$ $2x$ $+$ $65$ $\,=\,$ $25$ $+$ $x^2$ $-$ $10x$

$\implies$ $x^2$ $-$ $2x$ $+$ $65$ $-$ $x^2$ $+$ $10x$ $-$ $25$ $\,=\,$ $0$

$\implies$ $x^2$ $-$ $x^2$ $+$ $10x$ $-$ $2x$ $+$ $65$ $-$ $25$ $\,=\,$ $0$

Cancel the equal terms, add the like terms and finally subtract the numbers to complete the simplification of the expression on the left hand side of the equation.

$\implies$ $\cancel{x^2}$ $-$ $\cancel{x^2}$ $+$ $8x$ $+$ $40$ $\,=\,$ $0$

The given logarithmic equation is successfully simplified as a linear equation in one variable.

$\implies$ $8x$ $+$ $40$ $\,=\,$ $0$

Now, solve the linear equation in one variable to get the value of the variable $x$.

$\implies$ $8x$ $\,=\,$ $-40$

$\implies$ $x$ $\,=\,$ $\dfrac{-40}{8}$

$\implies$ $x$ $\,=\,$ $-\dfrac{40}{8}$

$\implies$ $x$ $\,=\,$ $-\dfrac{\cancel{40}}{\cancel{8}}$

$\,\,\,\therefore\,\,\,\,\,\,$ $x$ $\,=\,$ $-5$


In this problem, $\log_{5-x}{(x^2-2x+65)}$ $\,=\,$ $2$ is the given log equation and it is solved that $x = -5$. Now, substitute the value of $x$ on the left hand side of the equation.

$=\,\,\,$ $\log_{5-(-5)}{\big((-5)^2-2(-5)+65\big)}$

$=\,\,\,$ $\log_{10}{(25+10+65)}$

$=\,\,\,$ $\log_{10}{(100)}$

$=\,\,\,$ $\log_{10}{\big(10^2\big)}$

$=\,\,\,$ $2 \times \log_{10}{(10)}$

$=\,\,\,$ $2 \times 1$

$=\,\,\,$ $2$

It is evaluated that the value of $\log_{5-x}{(x^2-2x+65)}$ is equal to $2$ when the value of $x$ is equal to $-5$.

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