141 Solve LN equations

Bell:

1

Khloe is going to invest $7,100 and leave it in an account for 9 years. Assuming the interest is compounded continuously, what interest rate, to the nearest hundredth of a percent, would be required in order for Khloe to end up with $12,600?

Use y=a(e)^{rt}

Notes: Solving LN equations for the exact value

1

2\ln(2x+4)+10=4

1

4\ln(7x+4)+1=13

1

4\ln(7x+4)+1=13

Solving Log Equations with Rejected Answers!

1

\log_{3}{(2x^{2}+7)}-\log_{3}{(3x+6)}=1

1

\log_{2}{(x^{2}-6)}-\log_{2}{(3x+5)}=1

1

\log_{3}{(x-2)}+\log_{3}{(x+4)}=3

1

\log_{6}{(3x+5)}+\log_{6}{(3x-4)}=2

1

\log_{2}{(3x^{2}-2)}-\log_{2}{(4x+5)}=0

0

141 Solve LN equations

Bell:

Khloe is going to invest $7,100 and leave it in an account for 9 years. Assuming the interest is compounded continuously, what interest rate, to the nearest hundredth of a percent, would be required in order for Khloe to end up with $12,600?

Use y=a(e)^{rt}

Notes: Solving LN equations for the exact value

2\ln(2x+4)+10=4

4\ln(7x+4)+1=13

4\ln(7x+4)+1=13

Solving Log Equations with Rejected Answers!

\log_{3}{(2x^{2}+7)}-\log_{3}{(3x+6)}=1

\log_{2}{(x^{2}-6)}-\log_{2}{(3x+5)}=1

\log_{3}{(x-2)}+\log_{3}{(x+4)}=3

\log_{6}{(3x+5)}+\log_{6}{(3x-4)}=2

\log_{2}{(3x^{2}-2)}-\log_{2}{(4x+5)}=0

Choose all that are true

Bell:

Khloe is going to invest $7,100 and leave it in an account for 9 years. Assuming the interest is compounded continuously, what interest rate, to the nearest hundredth of a percent, would be required in order for Khloe to end up with $12,600?

Use y=a(e)^{rt}

Notes: Solving LN equations for the exact value

2\ln(2x+4)+10=4

4\ln(7x+4)+1=13

4\ln(7x+4)+1=13

Solving Log Equations with Rejected Answers!

\log_{3}{(2x^{2}+7)}-\log_{3}{(3x+6)}=1

\log_{2}{(x^{2}-6)}-\log_{2}{(3x+5)}=1

\log_{3}{(x-2)}+\log_{3}{(x+4)}=3

\log_{6}{(3x+5)}+\log_{6}{(3x-4)}=2

\log_{2}{(3x^{2}-2)}-\log_{2}{(4x+5)}=0

Choose all that are true

141 Solve LN equations

141 Solve LN equations

Bell:

Khloe is going to invest $7,100 and leave it in an account for 9 years. Assuming the interest is compounded continuously, what interest rate, to the nearest hundredth of a percent, would be required in order for Khloe to end up with $12,600?Use y=a(e)^{rt}

Notes: Solving LN equations for the exact value

2\ln(2x+4)+10=4

4\ln(7x+4)+1=13

4\ln(7x+4)+1=13

Solving Log Equations with Rejected Answers!

\log_{3}{(2x^{2}+7)}-\log_{3}{(3x+6)}=1

\log_{2}{(x^{2}-6)}-\log_{2}{(3x+5)}=1

\log_{3}{(x-2)}+\log_{3}{(x+4)}=3

\log_{6}{(3x+5)}+\log_{6}{(3x-4)}=2

\log_{2}{(3x^{2}-2)}-\log_{2}{(4x+5)}=0

Choose all that are true

Bell:

Khloe is going to invest $7,100 and leave it in an account for 9 years. Assuming the interest is compounded continuously, what interest rate, to the nearest hundredth of a percent, would be required in order for Khloe to end up with $12,600?Use y=a(e)^{rt}

Notes: Solving LN equations for the exact value

2\ln(2x+4)+10=4

4\ln(7x+4)+1=13

4\ln(7x+4)+1=13

Solving Log Equations with Rejected Answers!

\log_{3}{(2x^{2}+7)}-\log_{3}{(3x+6)}=1

\log_{2}{(x^{2}-6)}-\log_{2}{(3x+5)}=1

\log_{3}{(x-2)}+\log_{3}{(x+4)}=3

\log_{6}{(3x+5)}+\log_{6}{(3x-4)}=2

\log_{2}{(3x^{2}-2)}-\log_{2}{(4x+5)}=0

Choose all that are true

Khloe is going to invest $7,100 and leave it in an account for 9 years. Assuming the interest is compounded continuously, what interest rate, to the nearest hundredth of a percent, would be required in order for Khloe to end up with $12,600?

Use y=a(e)^{rt}

Khloe is going to invest $7,100 and leave it in an account for 9 years. Assuming the interest is compounded continuously, what interest rate, to the nearest hundredth of a percent, would be required in order for Khloe to end up with $12,600?

Use y=a(e)^{rt}