Logarithms and solving for x

[Louise Johnson]

I am completely stuck on two questions. The first is solving for x which I should be a pro at by now however it has me stumped. ( I have been trying to figure out how to solve for x using ln but nothing has clicked) I am not sure how important the second question really is as its a one off thing and I cannot find any other references in my books.

Thank you
Louise

Question #1

\(\displaystyle 4(7^{x - 1} ) = e^{x + 1}\)

Question #2

Write\(\displaystyle \log _a \sqrt 6\),in terms of p and Q, if\(\displaystyle \log _a 2 = p\)and\(\displaystyle \log _a 3 =Q\)

 

[pka]

\(\displaystyle \L\begin{array}{l}
4\left( {7^{x - 1} } \right) = e^{x + 1} \\
\left( {x - 1} \right)\left[ {\ln (7)} \right] = \left( {x + 1} \right) - \ln (4) \\
\left[ {\ln (7) - 1} \right]x = \ln \left( {\frac{7}{4}} \right) + 1 \\
\end{array}\)

Did you see this http://www.freemathhelp.com/forum/viewtopic.php?t=20937

 

[Louise Johnson]

Thank you PKA, Wow I can see why I didn't have a chance. It took me over ten minutes just to finish it off and plug in the value of x to check to see if I was correct! ( Yes I did spend some time with your other post but didn't really pay attention to the final direction of the added question)
Hopefully someone will have a go at the second one.
Thankyou again
Louise

 

[Mrspi]

Question #2

Write\(\displaystyle \log _a \sqrt 6\),in terms of p and Q, if\(\displaystyle \log _a 2 = p\)and\(\displaystyle \log _a 3 =Q\)

log<SUB>a</SUB> √6 can be written as log<SUB>a</SUB> 6<SUP>1/2</SUP>

Or, using one of the rules of logs,

(1/2) log<SUB>a</SUB> 6

(1/2) log<SUB>a</SUB> (2*3)

Use another one of the rules of logs:

(1/2) [log<SUB>a</SUB> 2 + log<SUB>a</SUB> 3 ]

You know that log<SUB>a</SUB> 2 = p, and that log<SUB>a</SUB> 3 = q. So, you can substitute and get this:

(1/2)(p + q)

 

[soroban]

Hello, Louise!

For the benefit of anyone else looking on . . .

Solve for \(\displaystyle x:\;\;4\cdot7^{x - 1} \:= \:e^{x + 1}\)

Take logs: \(\displaystyle \;\;\ln\left(4\cdot7^{x-1}\right)\;=\;\ln\left(e^{x-1}\right)\)

. . . . . .\(\displaystyle \ln(4)\,+\,\ln\left(7^{x-1}\right) \;=\;(x\,+\,1)\ln(e)\;\;\leftarrow\;\ln(e)\,=\,1\)

. . . \(\displaystyle \ln(4)\,+\,(x\,-\,1)\ln(7)\;=\;x\,+\,1\)

. .\(\displaystyle \ln(4)\,+\,x\ln(7)\,-\,\ln(7)\;=\;x\,+\,1\)

. . . . . . . . . \(\displaystyle x\ln(7)\,-\,x\;=\;\ln(7)\,-\,\ln(4)\,+\,1\)

. . . . . . . . \(\displaystyle x[\ln(7)\,-\,1]\;=\;\ln(7)\,-\,\ln(4)\,+\,1\)

. . . . . . . . . . . . . . . .\(\displaystyle x\;=\;\frac{\ln(7)\,-\,\ln(4)\,+\,1}{\ln(7)\,-\,1}\;\) **

2) Write \(\displaystyle \log _a \sqrt{6}\), in terms of \(\displaystyle P\) and \(\displaystyle Q\), if \(\displaystyle \log _a(2)\:=\\) and \(\displaystyle \log _a(3)\:=\:Q\)

We have: \(\displaystyle \:\log_a\left(6^{\frac{1}{2}}\right) \;=\;\frac{1}{2}\log_a(6)\;=\;\frac{1}{2}\log_a(2\cdot3)\)

. . \(\displaystyle = \;\frac{1}{2}[\underbrace{\log_a(2)}_{\downarrow}\,+\,\underbrace{\log_a(3)}_{\downarrow}]\)
. . \(\displaystyle = \;\frac{1}{2}\;[\;P\;\;\,+\;\;\;Q\;]\)

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

**

The first answer can be simplified beyond all recognition.

We have: \(\displaystyle \:\frac{\log(7)\,-\,\ln(4)\,+\,1}{\ln(7)\,-\,1}\)

Since \(\displaystyle \ln(e)\,=\,1\), it becomes: \(\displaystyle \:\frac{\ln(7)\,-\,\ln(4)\,+\,\ln(e)}{\ln(7)\,-\,\ln(e)}\)

. . and we have: \(\displaystyle \L\:\frac{\ln\left(\frac{7e}{4}\right)}{\ln\left(\frac{7}{e}\right)}\;\) or \(\displaystyle \L\;\log_{\frac{7}{e}}\left(\frac{7e}{4}\right)\)


Guarenteed to impress/surprise/terrify your teacher . . .

 

[Louise Johnson]

Thank you for all the posts. Its alot of work! I have been going through both questions again and the second one now seems simple in comparison. I can only hope I can get to the point where I can simplify logarithms like!!
Thank you
Sincerly
Louise