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Common Core: High School - Functions : Express Exponential Models as Logarithmic Solutions: CCSS.Math.Content.HSF-LE.A.4

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Example Questions

Example Question #1 : Express Exponential Models As Logarithmic Solutions: Ccss.Math.Content.Hsf Le.A.4

Solve for $\displaystyle x$ using rules of logarithmic functions.

$e^{x-4}+2=10$ $\displaystyle e^{x-4}+2=10$

Possible Answers:

$x=\frac{\ln 8}{2}$ $\displaystyle x=\frac{\ln 8}{2}$

$x=\ln 8+4$ $\displaystyle x=\ln 8+4$

$x={\ln 4}$ $\displaystyle x={\ln 4}$

$x=\ln 8-4$ $\displaystyle x=\ln 8-4$

$x=\frac{\ln 8}{4}$ $\displaystyle x=\frac{\ln 8}{4}$

Correct answer:

$x=\ln 8+4$ $\displaystyle x=\ln 8+4$

Explanation:

This question is testing one's ability to understand logarithmic rules and apply them in order to solve a function.

For the purpose of Common Core Standards, "For exponential models, express as a logarithm the solution to ab^(ct) = d where a, c, and dare numbers and the base b is 2, 10, or e; evaluate the logarithm using technology." falls within the Cluster A of "Construct and compare linear, quadratic, and exponential models and solve problems" concept (CCSS.MATH.CONTENT.HSF-LE.A.4).

Knowing the standard and the concept for which it relates to, we can now do the step-by-step process to solve the problem in question.

Step 1: Use algebraic operations to manipulate the function and isolate the $\displaystyle x$ value on one side of the equation.

$e^{x-4}+2=10$ $\displaystyle e^{x-4}+2=10$

Subtract two from both sides.

$\\e^{x-4}+2-2=10-2 \\e^{x-4}=8$ $\displaystyle \\e^{x-4}+2-2=10-2 \\e^{x-4}=8$

Step 2: Identify logarithmic rules.

Recall that $\ln e=1$ $\displaystyle \ln e=1$

Step 3: Apply logarithmic rules to solve for $\displaystyle x$.

$\\e^{x-4}=8 \\ \ln e^{x-4}=\ln 8\\ x-4=\ln 8\\x=\ln 8+4$ $\displaystyle \\e^{x-4}=8 \\ \ln e^{x-4}=\ln 8\\ x-4=\ln 8\\x=\ln 8+4$

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Example Question #1 : Express Exponential Models As Logarithmic Solutions: Ccss.Math.Content.Hsf Le.A.4

Solve for $\displaystyle x$ using rules of logarithmic functions.

$e^{x-4}+1=5$ $\displaystyle e^{x-4}+1=5$

Possible Answers:

$x=\ln4-4$ $\displaystyle x=\ln4-4$

$x=\ln4-1$ $\displaystyle x=\ln4-1$

$x=\ln1+4$ $\displaystyle x=\ln1+4$

$x=\ln4+4$ $\displaystyle x=\ln4+4$

$x=\ln4+1$ $\displaystyle x=\ln4+1$

Correct answer:

$x=\ln4+4$ $\displaystyle x=\ln4+4$

Explanation:

This question is testing one's ability to understand logarithmic rules and apply them in order to solve a function.

Knowing the standard and the concept for which it relates to, we can now do the step-by-step process to solve the problem in question.

Step 1: Use algebraic operations to manipulate the function and isolate the $\displaystyle x$ value on one side of the equation.

$e^{x-4}+1=5$ $\displaystyle e^{x-4}+1=5$

Subtract one from both sides.

$\\e^{x-4}+1-1=5-1 \\e^{x-4}=4$ $\displaystyle \\e^{x-4}+1-1=5-1 \\e^{x-4}=4$

Step 2: Identify logarithmic rules.

Recall that $\ln e=1$ $\displaystyle \ln e=1$

Step 3: Apply logarithmic rules to solve for $\displaystyle x$.

$\\e^{x-4}=4 \\ \ln e^{x-4}=\ln 4\\ x-4=\ln 4\\x=\ln 4+4$ $\displaystyle \\e^{x-4}=4 \\ \ln e^{x-4}=\ln 4\\ x-4=\ln 4\\x=\ln 4+4$

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Example Question #3 : Express Exponential Models As Logarithmic Solutions: Ccss.Math.Content.Hsf Le.A.4

Solve for $\displaystyle x$ using rules of logarithmic functions.

$2e^{x}+1=5$ $\displaystyle 2e^{x}+1=5$

Possible Answers:

$x=\ln1$ $\displaystyle x=\ln1$

$x=\ln2$ $\displaystyle x=\ln2$

$x=\sqrt{\ln4}$ $\displaystyle x=\sqrt{\ln4}$

$x=\ln4$ $\displaystyle x=\ln4$

$x=\sqrt{\ln2}$ $\displaystyle x=\sqrt{\ln2}$

Correct answer:

$x=\ln2$ $\displaystyle x=\ln2$

Explanation:

This question is testing one's ability to understand logarithmic rules and apply them in order to solve a function.

Knowing the standard and the concept for which it relates to, we can now do the step-by-step process to solve the problem in question.

Step 1: Use algebraic operations to manipulate the function and isolate the $\displaystyle x$ value on one side of the equation.

$2e^{x}+1=5$ $\displaystyle 2e^{x}+1=5$

Subtract one from both sides.

$\\2e^{x}+1-1=5-1 \\2e^{x}=4$ $\displaystyle \\2e^{x}+1-1=5-1 \\2e^{x}=4$

Step 2: Identify logarithmic rules.

Recall that $\ln e=1$ $\displaystyle \ln e=1$

Step 3: Apply logarithmic rules to solve for $\displaystyle x$.

$\\2e^x=4 \\e^x=\frac{4}{2} \\ \\e^x=2 \\ \ln e^x=\ln2 \\x=\ln 2$ $\displaystyle \\2e^x=4 \\e^x=\frac{4}{2} \\ \\e^x=2 \\ \ln e^x=\ln2 \\x=\ln 2$

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Example Question #2 : Express Exponential Models As Logarithmic Solutions: Ccss.Math.Content.Hsf Le.A.4

Solve for $\displaystyle x$ using rules of logarithmic functions.

$2e^{x}+5=17$ $\displaystyle 2e^{x}+5=17$

Possible Answers:

$x=\ln6+2$ $\displaystyle x=\ln6+2$

$x=\ln4-2$ $\displaystyle x=\ln4-2$

$x=\ln6$ $\displaystyle x=\ln6$

$x=\ln4$ $\displaystyle x=\ln4$

$x=\ln12-2$ $\displaystyle x=\ln12-2$

Correct answer:

$x=\ln6$ $\displaystyle x=\ln6$

Explanation:

This question is testing one's ability to understand logarithmic rules and apply them in order to solve a function.

Knowing the standard and the concept for which it relates to, we can now do the step-by-step process to solve the problem in question.

Step 1: Use algebraic operations to manipulate the function and isolate the $\displaystyle x$ value on one side of the equation.

$2e^{x}+5=17$ $\displaystyle 2e^{x}+5=17$

Subtract five from both sides and then divide by two.

$\\2e^{x}+5-5=17-5\\2e^{x}=12\\e^x=6$ $\displaystyle \\2e^{x}+5-5=17-5\\2e^{x}=12\\e^x=6$

Step 2: Identify logarithmic rules.

Recall that $\ln e=1$ $\displaystyle \ln e=1$

Step 3: Apply logarithmic rules to solve for $\displaystyle x$.

$\\e^x=6 \\ \ln e^x=\ln6 \\x=\ln 6$ $\displaystyle \\e^x=6 \\ \ln e^x=\ln6 \\x=\ln 6$

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Example Question #1 : Express Exponential Models As Logarithmic Solutions: Ccss.Math.Content.Hsf Le.A.4

Solve for $\displaystyle x$ using rules of logarithmic functions.

$4e^{x}+5=17$ $\displaystyle 4e^{x}+5=17$

Possible Answers:

$x=\ln6$ $\displaystyle x=\ln6$

x=1 $\displaystyle x=1$

x=3 $\displaystyle x=3$

$x=-\ln3$ $\displaystyle x=-\ln3$

$x=\ln3$ $\displaystyle x=\ln3$

Correct answer:

$x=\ln3$ $\displaystyle x=\ln3$

Explanation:

This question is testing one's ability to understand logarithmic rules and apply them in order to solve a function.

Knowing the standard and the concept for which it relates to, we can now do the step-by-step process to solve the problem in question.

Step 1: Use algebraic operations to manipulate the function and isolate the $\displaystyle x$ value on one side of the equation.

$4e^{x}+5=17$ $\displaystyle 4e^{x}+5=17$

Subtract five from both sides and then divide by four.

$\\4e^{x}+5-5=17-5\\4e^{x}=12\\e^x=3$ $\displaystyle \\4e^{x}+5-5=17-5\\4e^{x}=12\\e^x=3$

Step 2: Identify logarithmic rules.

Recall that $\ln e=1$ $\displaystyle \ln e=1$

Step 3: Apply logarithmic rules to solve for $\displaystyle x$.

$\\e^x=3 \\ \ln e^x=\ln3 \\x=\ln 3$ $\displaystyle \\e^x=3 \\ \ln e^x=\ln3 \\x=\ln 3$

Report an Error

Example Question #6 : Express Exponential Models As Logarithmic Solutions: Ccss.Math.Content.Hsf Le.A.4

Solve for $\displaystyle x$ using rules of logarithmic functions.

$4e^{x}+5=13$ $\displaystyle 4e^{x}+5=13$

Possible Answers:

$x=\ln2-1$ $\displaystyle x=\ln2-1$

$x=\ln1$ $\displaystyle x=\ln1$

$x=\ln2$ $\displaystyle x=\ln2$

x=2 $\displaystyle x=2$

x=e^2 $\displaystyle x=e^2$

Correct answer:

$x=\ln2$ $\displaystyle x=\ln2$

Explanation:

This question is testing one's ability to understand logarithmic rules and apply them in order to solve a function.

Knowing the standard and the concept for which it relates to, we can now do the step-by-step process to solve the problem in question.

Step 1: Use algebraic operations to manipulate the function and isolate the $\displaystyle x$ value on one side of the equation.

$4e^{x}+5=13$ $\displaystyle 4e^{x}+5=13$

Subtract five from both sides and then divide by four.

$\\4e^{x}+5-5=13-5\\4e^{x}=8\\e^x=2$ $\displaystyle \\4e^{x}+5-5=13-5\\4e^{x}=8\\e^x=2$

Step 2: Identify logarithmic rules.

Recall that $\ln e=1$ $\displaystyle \ln e=1$

Step 3: Apply logarithmic rules to solve for $\displaystyle x$.

$\\e^x=2 \\ \ln e^x=\ln2 \\x=\ln 2$ $\displaystyle \\e^x=2 \\ \ln e^x=\ln2 \\x=\ln 2$

Report an Error

Example Question #5 : Express Exponential Models As Logarithmic Solutions: Ccss.Math.Content.Hsf Le.A.4

Solve for $\displaystyle x$ using rules of logarithmic functions.

$4e^{x}-6=12$ $\displaystyle 4e^{x}-6=12$

Possible Answers:

$x=\ln \frac{9}{2}$ $\displaystyle x=\ln \frac{9}{2}$

$x=\ln\frac{2}{9}$ $\displaystyle x=\ln\frac{2}{9}$

$x=-\ln18$ $\displaystyle x=-\ln18$

$x=\ln2+18$ $\displaystyle x=\ln2+18$

$x=\ln18-2$ $\displaystyle x=\ln18-2$

Correct answer:

$x=\ln \frac{9}{2}$ $\displaystyle x=\ln \frac{9}{2}$

Explanation:

This question is testing one's ability to understand logarithmic rules and apply them in order to solve a function.

Knowing the standard and the concept for which it relates to, we can now do the step-by-step process to solve the problem in question.

Step 1: Use algebraic operations to manipulate the function and isolate the $\displaystyle x$ value on one side of the equation.

$4e^{x}-6=12$ $\displaystyle 4e^{x}-6=12$

Add six from both sides and then divide by four.

$\\4e^{x}-6+6=12+6\\4e^{x}=18\\\\e^x=\frac{9}{2}$ $\displaystyle \\4e^{x}-6+6=12+6\\4e^{x}=18\\\\e^x=\frac{9}{2}$

Step 2: Identify logarithmic rules.

Recall that $\ln e=1$ $\displaystyle \ln e=1$

Step 3: Apply logarithmic rules to solve for $\displaystyle x$.

$\\e^x=\frac{9}{2} \\\\ \ln e^x=\ln\frac{9}{2} \\x=\ln \frac{9}{2}$ $\displaystyle \\e^x=\frac{9}{2} \\\\ \ln e^x=\ln\frac{9}{2} \\x=\ln \frac{9}{2}$

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Example Question #1 : Express Exponential Models As Logarithmic Solutions: Ccss.Math.Content.Hsf Le.A.4

Solve for $\displaystyle x$ using rules of logarithmic functions.

$e^{x-4}-2=10$ $\displaystyle e^{x-4}-2=10$

Possible Answers:

$x=\ln6-2$ $\displaystyle x=\ln6-2$

$x=\ln4$ $\displaystyle x=\ln4$

$x=\ln12+4$ $\displaystyle x=\ln12+4$

$x=\ln6+2$ $\displaystyle x=\ln6+2$

$x=\ln12-4$ $\displaystyle x=\ln12-4$

Correct answer:

$x=\ln12+4$ $\displaystyle x=\ln12+4$

Explanation:

This question is testing one's ability to understand logarithmic rules and apply them in order to solve a function.

Knowing the standard and the concept for which it relates to, we can now do the step-by-step process to solve the problem in question.

Step 1: Use algebraic operations to manipulate the function and isolate the $\displaystyle x$ value on one side of the equation.

$e^{x-4}-2=10$ $\displaystyle e^{x-4}-2=10$

Add two from both sides.

$\\e^{x-4}-2+2=10+2 \\e^{x-4}=12$ $\displaystyle \\e^{x-4}-2+2=10+2 \\e^{x-4}=12$

Step 2: Identify logarithmic rules.

Recall that $\ln e=1$ $\displaystyle \ln e=1$

Step 3: Apply logarithmic rules to solve for $\displaystyle x$.

$\\e^{x-4}=12 \\ \ln e^{x-4}=\ln 12\\ x-4=\ln 12\\x=\ln 12+4$ $\displaystyle \\e^{x-4}=12 \\ \ln e^{x-4}=\ln 12\\ x-4=\ln 12\\x=\ln 12+4$

Report an Error

Example Question #1 : Express Exponential Models As Logarithmic Solutions: Ccss.Math.Content.Hsf Le.A.4

Solve for $\displaystyle x$ using rules of logarithmic functions.

$e^{x-4}-1=5$ $\displaystyle e^{x-4}-1=5$

Possible Answers:

$x=\ln 6-4$ $\displaystyle x=\ln 6-4$

$x=\ln 6-2$ $\displaystyle x=\ln 6-2$

$x=\ln 6+2$ $\displaystyle x=\ln 6+2$

$x=\ln 6$ $\displaystyle x=\ln 6$

$x=\ln 6+4$ $\displaystyle x=\ln 6+4$

Correct answer:

$x=\ln 6+4$ $\displaystyle x=\ln 6+4$

Explanation:

This question is testing one's ability to understand logarithmic rules and apply them in order to solve a function.

Knowing the standard and the concept for which it relates to, we can now do the step-by-step process to solve the problem in question.

Step 1: Use algebraic operations to manipulate the function and isolate the $\displaystyle x$ value on one side of the equation.

$e^{x-4}-1=5$ $\displaystyle e^{x-4}-1=5$

Add one from both sides.

$\\e^{x-4}-1+1=5+1 \\e^{x-4}=6$ $\displaystyle \\e^{x-4}-1+1=5+1 \\e^{x-4}=6$

Step 2: Identify logarithmic rules.

Recall that $\ln e=1$ $\displaystyle \ln e=1$

Step 3: Apply logarithmic rules to solve for $\displaystyle x$.

$\\e^{x-4}=6 \\ \ln e^{x-4}=\ln 6\\ x-4=\ln 6\\x=\ln 6+4$ $\displaystyle \\e^{x-4}=6 \\ \ln e^{x-4}=\ln 6\\ x-4=\ln 6\\x=\ln 6+4$

Report an Error

Example Question #10 : Express Exponential Models As Logarithmic Solutions: Ccss.Math.Content.Hsf Le.A.4

Solve for $\displaystyle x$ using rules of logarithmic functions.

$e^{x-4}-1=1$ $\displaystyle e^{x-4}-1=1$

Possible Answers:

$x=\ln 2-4$ $\displaystyle x=\ln 2-4$

$x=\ln 2+2$ $\displaystyle x=\ln 2+2$

$x=\ln 2+4$ $\displaystyle x=\ln 2+4$

$x=\ln 2-2$ $\displaystyle x=\ln 2-2$

$x=\ln 2+1$ $\displaystyle x=\ln 2+1$

Correct answer:

$x=\ln 2+4$ $\displaystyle x=\ln 2+4$

Explanation:

This question is testing one's ability to understand logarithmic rules and apply them in order to solve a function.

Knowing the standard and the concept for which it relates to, we can now do the step-by-step process to solve the problem in question.

Step 1: Use algebraic operations to manipulate the function and isolate the $\displaystyle x$ value on one side of the equation.

$e^{x-4}-1=1$ $\displaystyle e^{x-4}-1=1$

Add one from both sides.

$\\e^{x-4}-1+1=1+1 \\e^{x-4}=2$ $\displaystyle \\e^{x-4}-1+1=1+1 \\e^{x-4}=2$

Step 2: Identify logarithmic rules.

Recall that $\ln e=1$ $\displaystyle \ln e=1$

Step 3: Apply logarithmic rules to solve for $\displaystyle x$.

$\\e^{x-4}=2 \\ \ln e^{x-4}=\ln 2\\ x-4=\ln 2\\x=\ln 2+4$ $\displaystyle \\e^{x-4}=2 \\ \ln e^{x-4}=\ln 2\\ x-4=\ln 2\\x=\ln 2+4$

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Common Core: High School - Functions : Express Exponential Models as Logarithmic Solutions: CCSS.Math.Content.HSF-LE.A.4

Study concepts, example questions & explanations for Common Core: High School - Functions

All Common Core: High School - Functions Resources

Example Questions

Example Question #1 : Express Exponential Models As Logarithmic Solutions: Ccss.Math.Content.Hsf Le.A.4

Example Question #1 : Express Exponential Models As Logarithmic Solutions: Ccss.Math.Content.Hsf Le.A.4

Example Question #3 : Express Exponential Models As Logarithmic Solutions: Ccss.Math.Content.Hsf Le.A.4

Example Question #2 : Express Exponential Models As Logarithmic Solutions: Ccss.Math.Content.Hsf Le.A.4

Example Question #1 : Express Exponential Models As Logarithmic Solutions: Ccss.Math.Content.Hsf Le.A.4

Example Question #6 : Express Exponential Models As Logarithmic Solutions: Ccss.Math.Content.Hsf Le.A.4

Example Question #5 : Express Exponential Models As Logarithmic Solutions: Ccss.Math.Content.Hsf Le.A.4

Example Question #1 : Express Exponential Models As Logarithmic Solutions: Ccss.Math.Content.Hsf Le.A.4

Example Question #1 : Express Exponential Models As Logarithmic Solutions: Ccss.Math.Content.Hsf Le.A.4

Example Question #10 : Express Exponential Models As Logarithmic Solutions: Ccss.Math.Content.Hsf Le.A.4

All Common Core: High School - Functions Resources

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