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$y=\log_{2}(4x^{2}-5x+1)$

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Convert between exponential and logarithmic form: all bases

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Q.

y=\log_{2}(4x^{2}-5x+1)

Identify key elements: Identify the base $b$ , the argument $x$ , and the result $y$ in the logarithmic equation. $\newline$ The equation is in the form $\log_b(x) = y$ , where $b$ is the base of the logarithm, $x$ is the argument, and $y$ is the result. $\newline$ For $y=\log_2(4x^2-5x+1)$ , the base $b$ is $2$ , the argument $x$ is $x$ $1$ , and the result $y$ is $y$ .
Convert to exponential form: Convert the logarithmic equation to its exponential form. $\newline$ The exponential form of a logarithmic equation $\log_b(x) = y$ is $b^y = x$ . $\newline$ Substitute $b=2$ , $y=y$ , and $x=(4x^2-5x+1)$ into the exponential form. $\newline$ Exponential equation: $2^y = 4x^2-5x+1$
Check for correctness: Check the exponential equation to ensure it is correctly formed and matches the original logarithmic equation. $\newline$ The original logarithmic equation was $y=\log_2(4x^2-5x+1)$ , and the exponential form is $2^y = 4x^2-5x+1$ . $\newline$ Both equations represent the same relationship between $y$ and the expression $(4x^2-5x+1)$ , so the conversion is correct.

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