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Algebra 4 Online
OpenStudy (anonymous):

Use natural logarithms to solve the equation. 6e2x – 9 = 23 Round to the nearest thousandth.

OpenStudy (anonymous):

\[6e^{2x}-9=23\]?

OpenStudy (anonymous):

yes

OpenStudy (anonymous):

1) add 9

OpenStudy (anonymous):

2) divide by 6

OpenStudy (anonymous):

let me know when you get \[e^{2x}=\frac{16}{3}\]

OpenStudy (anonymous):

e^2 x - 16/3 = 0

OpenStudy (anonymous):

still wating

OpenStudy (anonymous):

fan me so I can ask you a question

OpenStudy (anonymous):

satellite73 can you answer the most recent question

OpenStudy (calculusfunctions):

From the step @satellite73 left off\[e ^{2x}=\frac{ 16 }{ 3 }\]Take the natural logarithm of both sides\[\ln e ^{2x}=\ln \frac{ 16 }{ 3 }\]Now use the appropriate properties of logarithms to simplify and solve for x. Remember\[\ln e =1\]

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