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log_(2)128=

log⁡2128= \log _{2} 128=

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Full solution

Q. log⁡2128= \log _{2} 128=
  1. Identify base and argument: Identify the base (b) and the argument (x) of the logarithm.\newlineIn the expression log⁡2(128)\log_2(128), the base bb is 22 and the argument xx is 128128.
  2. Recall logarithm definition: Recall the definition of a logarithm.\newlineThe logarithmic equation log⁡2(x)=y\log_2(x) = y is equivalent to the exponential equation 2y=x2^y = x.
  3. Find exponent for equation: Find the exponent (yy) that makes the equation 2y=1282^y = 128 true.\newlineWe know that 27=1282^7 = 128 because 2×2×2×2×2×2×2=1282 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 = 128.
  4. Write equation in exponential form: Write the logarithmic equation in exponential form.\newlineSince 27=1282^7 = 128, we can write log⁡2(128)=7\log_2(128) = 7.

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