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Question 29 of 40Solve for x. 4-25× = 21.OA. 0.48OB. 0.31OC. 0.42OD. 0.08NOUSSA

Question 29 of 40Solve for x. 4-25× = 21.OA. 0.48OB. 0.31OC. 0.42OD. 0.08NOUSSA-example-1
User Tjbp
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ANSWER


A.\text{ }0.48

Step-by-step explanation

We want to solve for x in the equation:


4*2^(5x)=21

First, divide both sides of the equation by 4:


\begin{gathered} (4*2^(5x))/(4)=(21)/(4) \\ \\ 2^(5x)=5.25 \end{gathered}

Find the logarithm of both sides of the equation:


\log2^(5x)=\log5.25

Simplify the left-hand side of the equation:


5x(\log2)=\log5.25

Divide both sides by log2:


5x=(\log5.25)/(\log2)=2.392

Divide both sides of the equation by 5:


\begin{gathered} (5x)/(5)=(2.392)/(5) \\ \\ x=0.48 \end{gathered}

The answer is option A.

User Nick Schmidt
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