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Mike is working on solving the exponential equation 74^x = 37; however, he is not quite sure where to start. Using complete sentences, describe to Mike how to solve this equation.

User Dib
by
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2 Answers

9 votes

Answer:


x = \log_(74)(37)

Explanation:

we want to solve the following exponential equation:


\iff {74}^(x) = 37

Taking common logarithm of both sides yields:


\iff \log{74}^(x) = \log37

recall that,


  • log( {a}^(x) ) = x log(a)

utilizing it yields:


\iff x \cdot\log{74} = \log37

divide both sides by log74:


\iff \frac{ x \cdot\log{74}}{ \log(74)} = ( \log37)/( \log(74) )

remember that,


log_b( {a} ) = (log(a))/(log(b))

hence,


x = \boxed{ \log_(74)(37)}

User Klutt
by
7.7k points
8 votes

Answer:

Method 1


74^x = 37

Take natural logs of both sides:


\implies \ln74^x = \ln37

Apply the log power rule:
\ln(a)^b=b \ln (a)


\implies x\ln74 = \ln37

Divide both sides by ln 74:


\implies (x\ln74)/(\ln74) = (\ln37)/(\ln74)


\implies x= (\ln37)/(\ln74)


\implies x=0.8389552282...

Method 1


74^x = 37

Take logs of base 74 of both sides:


\implies \log_(74)74^x = \log_(74)37

Apply the log power rule:
\log_am^n=n\log_am


\implies x\log_(74)74= \log_(74)37

Apply log of the same number as base rule:
\log_aa=1


\implies x= \log_(74)37


\implies x=0.8389552282...

User Revolutionkpi
by
8.5k points

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