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zepelin [54]
4 years ago
10

Given the exponential equation 2x = 128, what is the logarithmic form of the equation in base 10?

Mathematics
2 answers:
DIA [1.3K]4 years ago
7 0
2^{x} = 128
log_2 128 = x
x = \frac{log_{10}128}{log_{10}2}
max2010maxim [7]4 years ago
5 0

Answer:

log_{2}128= \frac{log_{10} 128}{log_{10} 2}.  

Step-by-step explanation:

Given : 2^{x} = 128.

To find :  what is the logarithmic form of the equation in base 10

Solution : We have given that 2^{x} = 128.

By th change base rule ( inverse of exponential function )

a^{b} = c is equal to  log_{a}(c) = b

Then 2^{x} = 128 in to logarithm form  log_{2}(128) = x.

Then in to the base 10 logarithmic form.

By the change of base formula log_{b}a= \frac{log_{10} a}{log_{10} b}.

Then , log_{2}128= \frac{log_{10} 128}{log_{10} 2}.  

Therefore, log_{2}128= \frac{log_{10} 128}{log_{10} 2}.  

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