How do you write the equation $7^{- 2} = \frac{1}{49}$ $7^-2=1/49$ in logarithmic form?

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How do you write the equation $7^{- 2} = \frac{1}{49}$ $7^-2=1/49$ in logarithmic form?

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Answer 1 · 2018-05-10T06:48:35

${log}_{7} (\frac{1}{49}) = - 2$ $log_7(1/49)=-2$

Explanation:

If $b^{x} = y$ $b^x=y$ , then ${log}_{b} y = x$ $log_by=x$
Therefore, since $7^{- 2} = \frac{1}{49}$ $7^-2=1/49$ , then ${log}_{7} (\frac{1}{49}) = - 2$ $log_7(1/49)=-2$

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How do you write the equation $7^{- 2} = \frac{1}{49}$ $7^-2=1/49$ in logarithmic form?

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How do you write the equation $7^{- 2} = \frac{1}{49}$ $7^-2=1/49$ in logarithmic form?