Solve the equation.
[tex]\begin{array}{l}
2^{x-1}=31 \\
x=[?]\end{array}[/tex]
Answer (2)
Take the logarithm base 2 of both sides: lo g 2 ( 2 x − 1 ) = lo g 2 ( 31 ) .
Simplify using logarithm properties: x − 1 = lo g 2 ( 31 ) .
Isolate x : x = lo g 2 ( 31 ) + 1 .
The solution is approximately: x = l n ( 2 ) l n ( 31 ) + 1 ≈ 5.954 . Therefore, x = 5.954
Explanation
Problem Analysis We are given the equation 2 x − 1 = 31 and we want to solve for x .
Applying Logarithms To solve for x , we can take the logarithm base 2 of both sides of the equation. This gives us: lo g 2 ( 2 x − 1 ) = lo g 2 ( 31 ) Using the property of logarithms that lo g b ( b a ) = a , we can simplify the left side of the equation: x − 1 = lo g 2 ( 31 )
Isolating x Now, we isolate x by adding 1 to both sides of the equation: x = lo g 2 ( 31 ) + 1
Change of Base We can use the change of base formula to express the logarithm in terms of the natural logarithm (ln): lo g 2 ( 31 ) = ln ( 2 ) ln ( 31 ) So, we have: x = ln ( 2 ) ln ( 31 ) + 1
Calculating the Value Using a calculator, we find that l n ( 2 ) l n ( 31 ) ≈ 4.954 . Therefore, x ≈ 4.954 + 1 = 5.954
Final Answer Thus, the solution for x is approximately 5.954 .
Examples
Exponential equations like this one are used in various fields, such as calculating the growth of bacteria, determining the decay of radioactive substances, or modeling compound interest in finance. For instance, if you invest money in an account that compounds annually, you can use exponential equations to determine how long it will take for your investment to reach a certain value. Understanding exponential equations helps in making informed decisions about investments and predicting future growth.
To solve the equation 2 x − 1 = 31 , we use logarithms to express it as x = lo g 2 ( 31 ) + 1 , which we can calculate using the change of base formula. The final approximate value of x is 5.954 .
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