Question

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

Answer 1

@Michele_Laino i wanna say 0 because http://www.wolframalpha.com/input/?i=7e%5E3%5Ex+%3D+312

Answer 2

or maybe C

Answer 3

I think that we can divide both sides by \(7\), so we can write:
\[\Huge {e^{{3^x}}} \cong 45\]

Answer 4

oh okay

Answer 5

subsequently, we can take the natural logaritm, like this:
\[\Huge {3^x} \cong \ln 45\]

Answer 6

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

Answer 7

Seems good to me.

Answer 8

thanks!! :) @Jhannybean

Answer 9

Okay so do we solve that or do nother step?

Answer 10

it is simplewe have to evaluate this quantity:
\[\Huge x \cong {\log _3}\left( {\ln 45} \right) = ...?\]

Answer 11

i dont know how

Answer 12

first step:
using a scientific calculator, we can write this:
\[\Huge \ln 45 \cong 3.81\]

Answer 13

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
\[\huge {\log _3}\left( {\ln 45} \right) = \frac{{{{\log }_{10}}\left( {\ln 45} \right)}}{{{{\log }_{10}}3}} = ...?\]

Answer 14

2.69

Answer 15

more steps:
\[\huge \begin{gathered}
{\log _3}\left( {\ln 45} \right) = \frac{{{{\log }_{10}}\left( {\ln 45} \right)}}{{{{\log }_{10}}3}} = \hfill \\
\hfill \\
= \frac{{{{\log }_{10}}3.81}}{{{{\log }_{10}}3}} = ...? \hfill \\
\end{gathered} \]

Answer 16

I got 1.22

Answer 17

thats not an answer choice though?

Answer 18

I think that we can pick option B

Answer 19

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.

Answer 20

here we have arrived to this result:
\[\huge {\log _2}x + {\log _2}\left( {x - 7} \right) = {\log _2}8\]
right?

Answer 21

right

Answer 22

now I can apply this property of logarithms: \(\log m +\log n=\log (m \cdot n)\)
so I can write this:
\[\huge {\log _2}\left\{ {x \cdot \left( {x + 7} \right)} \right\} = {\log _2}8\]

Answer 23

oh okay

Answer 24

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?

Answer 25

oops... So*

Answer 26

B?

Answer 27

no, since we have not the logaritms now, please we have a quadratic equation which we have to solve

Answer 28

D

Answer 29

that's right!

Answer 30

Thank you so much!!

Answer 31

:)