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Write the log equation as an exponential equation. You do not need to solve for x. log_(5)(x^(2)-2x+20)=5x Answer:

Write the log equation as an exponential equation. You do not need to solve for x \mathrm{x} .\newlinelog⁡5(x2−2x+20)=5x \log _{5}\left(x^{2}-2 x+20\right)=5 x \newlineAnswer:

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Q. Write the log equation as an exponential equation. You do not need to solve for x \mathrm{x} .\newlinelog⁡5(x2−2x+20)=5x \log _{5}\left(x^{2}-2 x+20\right)=5 x \newlineAnswer:
  1. Identify components: Identify the base bb, the argument xx, and the exponent yy in the logarithmic equation.\newlineThe logarithmic equation is given as log⁡5(x2−2x+20)=5x\log_{5}(x^{2}-2x+20)=5x. Here, the base bb is 55, the argument xx is x2−2x+20x^{2}-2x+20, and the exponent yy is 5x5x.
  2. Convert to exponential form: Convert the logarithmic equation to its equivalent exponential form.\newlineUsing the definition of a logarithm, we can convert the equation log⁡5(x2−2x+20)=5x\log_{5}(x^{2}-2x+20)=5x to its exponential form by raising the base to the power of the exponent to get the argument. This gives us 55x=x2−2x+205^{5x} = x^{2}-2x+20.
  3. Write corresponding equation: Write down the exponential equation that corresponds to the given logarithmic equation.\newlineThe exponential equation that corresponds to log⁡5(x2−2x+20)=5x\log_{5}(x^{2}-2x+20)=5x is 55x=x2−2x+205^{5x} = x^{2}-2x+20.

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