anonymous
  • anonymous
Use the function: 7e^3^x = 312 The approximate solution would be: Select one: a. 0 b. 1.3 c. 1.99 d. 2.4
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
@Michele_Laino i wanna say 0 because http://www.wolframalpha.com/input/?i=7e%5E3%5Ex+%3D+312
anonymous
  • anonymous
or maybe C
Michele_Laino
  • Michele_Laino
I think that we can divide both sides by \(7\), so we can write: \[\Huge {e^{{3^x}}} \cong 45\]

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
oh okay
Michele_Laino
  • Michele_Laino
subsequently, we can take the natural logaritm, like this: \[\Huge {3^x} \cong \ln 45\]
Michele_Laino
  • Michele_Laino
and finally, we can take the logarithm with base 3: \[\Huge x \cong {\log _3}\left( {\ln 46} \right)\] what do you think about procedure? @Jhannybean
Jhannybean
  • Jhannybean
Seems good to me.
Michele_Laino
  • Michele_Laino
thanks!! :) @Jhannybean
anonymous
  • anonymous
Okay so do we solve that or do nother step?
Michele_Laino
  • Michele_Laino
it is simplewe have to evaluate this quantity: \[\Huge x \cong {\log _3}\left( {\ln 45} \right) = ...?\]
anonymous
  • anonymous
i dont know how
Michele_Laino
  • Michele_Laino
first step: using a scientific calculator, we can write this: \[\Huge \ln 45 \cong 3.81\]
Michele_Laino
  • Michele_Laino
second step: In order to evaluate the \(\log_3\) we have to change the basis of the logaritm, namely from 3 to 10, using this identity \[\huge {\log _3}\left( {\ln 45} \right) = \frac{{{{\log }_{10}}\left( {\ln 45} \right)}}{{{{\log }_{10}}3}} = ...?\]
anonymous
  • anonymous
2.69
Michele_Laino
  • Michele_Laino
more steps: \[\huge \begin{gathered} {\log _3}\left( {\ln 45} \right) = \frac{{{{\log }_{10}}\left( {\ln 45} \right)}}{{{{\log }_{10}}3}} = \hfill \\ \hfill \\ = \frac{{{{\log }_{10}}3.81}}{{{{\log }_{10}}3}} = ...? \hfill \\ \end{gathered} \]
Michele_Laino
  • Michele_Laino
I got 1.22
anonymous
  • anonymous
thats not an answer choice though?
Michele_Laino
  • Michele_Laino
I think that we can pick option B
anonymous
  • anonymous
okay great Use the following equation: log2x + log2(x – 7) = 3 The third step would be to: Select one: a. Graph the function. b. Take the log of each side. c. Square the expression. d. Factor the resulting quadratic equation.
Michele_Laino
  • Michele_Laino
here we have arrived to this result: \[\huge {\log _2}x + {\log _2}\left( {x - 7} \right) = {\log _2}8\] right?
anonymous
  • anonymous
right
Michele_Laino
  • Michele_Laino
now I can apply this property of logarithms: \(\log m +\log n=\log (m \cdot n)\) so I can write this: \[\huge {\log _2}\left\{ {x \cdot \left( {x + 7} \right)} \right\} = {\log _2}8\]
anonymous
  • anonymous
oh okay
Michele_Laino
  • Michele_Laino
therefore, I equate both numbers of such logarithms, so I get: \[\huge \begin{gathered} x \cdot \left( {x + 7} \right) = 8 \hfill \\ \hfill \\ x \cdot \left( {x + 7} \right) - 8 = 0 \hfill \\ \end{gathered} \] and the last expression is a quadratic equation. SO, what is the right option?
Michele_Laino
  • Michele_Laino
oops... So*
anonymous
  • anonymous
B?
Michele_Laino
  • Michele_Laino
no, since we have not the logaritms now, please we have a quadratic equation which we have to solve
anonymous
  • anonymous
D
Michele_Laino
  • Michele_Laino
that's right!
anonymous
  • anonymous
Thank you so much!!
Michele_Laino
  • Michele_Laino
:)

Looking for something else?

Not the answer you are looking for? Search for more explanations.