# Logarithms

• Jan 2nd 2010, 05:12 PM
ntrantrinh
Logarithms
Hi everyone,

I've been given about 15 questions to work on over the holiday and I'm still in the progress of solving them. But there are a few I really cannot solve.

1. $\displaystyle log_3 8 = p$ and $\displaystyle log_3 5 = q$, express $\displaystyle log 5$ in therms of p and q
2. If $\displaystyle a = log_7 (11 - 6 \sqrt{2})$ and $\displaystyle b = log_7 (45 + 29 \sqrt{2})$, find $\displaystyle 3a + 2b$ in simplest form
3. Solve the equation $\displaystyle log_2 x + log_4 x + log_8 x = 11$

• Jan 2nd 2010, 05:57 PM
ANDS!
#1 I'm not sure about so I'll leave that to someone else. Some trick that is escaping me at the moment.

#2 Remember:

$\displaystyle \log_{a}x^r=r\log_{a}x$ Apply the appropriate exponents, multiply out and use your properties of addition for logarithms to further simplify.

#3 Is where we want to use the change of base rule:

$\displaystyle \log_{a}b=\frac{\log_{c}b}{\log_{c}a}$

In this instance, we want to change the log base 4 and log base 8 into something that we can merge with our log base 2. Seeing as 4 and 8 are powers of 2, this seems like an appropriate way to go:

$\displaystyle \log_{2}x+\frac{\log_{2}x}{\log_{2}4}+\frac{\log_{ 2}x}{\log_{2}8}=11$
$\displaystyle \log_{2}x+\frac{\log_{2}x}{2}+\frac{\log_{2}x}{3}= 11$

$\displaystyle \frac{6\log_{2}x+3\log_{2}x+2\log_{2}x}{6}=11$

$\displaystyle 11\log_{2}x=66$

$\displaystyle \log_{2}x=6$

$\displaystyle 2^6=x$

$\displaystyle x=64$
• Jan 2nd 2010, 06:33 PM
Sudharaka
Dear ntrantrinh,

In the file attached below I have answered the first part. Please don't hesitate to reply me if you find the attachment unreadable, or if you have any further questions.

By the way we could try the second part if you would tell us what do you mean by sqrt[29]2

Hope this helps.
• Jan 2nd 2010, 07:18 PM
ntrantrinh
Quote:

Originally Posted by Sudharaka
Dear ntrantrinh,

In the file attached below I have answered the first part. Please don't hesitate to reply me if you find the attachment unreadable, or if you have any further questions.

By the way we could try the second part if you would tell us what do you mean by sqrt[29]2

Hope this helps.

I meant $\displaystyle \sqrt[29]{2}$ sorry i'll correct that now.
• Jan 2nd 2010, 10:03 PM
Soroban
Hello, ntrantrinh!

Here's my approach to #3 . . .

Quote:

3. Solve the equation: .$\displaystyle \log_2 x + \log_4 x + \log_8 x \:=\: 11$ .[1]

Let: .$\displaystyle \log_2x \,=\,P \quad\Rightarrow\quad x\:=\:2^P$ .[2]

Let: .$\displaystyle \log_4x \,=\,Q \quad\Rightarrow\quad x\:=\:4^Q \:=\:(2^2)^Q \:=\:2^{2Q}$ .[3]

Let: .$\displaystyle \log_8x \,=\,R \quad\Rightarrow\quad x \:=\:8^R \:=\:(2^3)^R \:=\:2^{3R}$ .[4]

Equate [2] and [4]: .$\displaystyle 2^P \:=\:2^{3R} \quad\Rightarrow\quad P \:=\:3R \quad\Rightarrow\quad \log_2x \:=\:3\log_8x$

Equate [3] and [4]: .$\displaystyle 2^{2Q} \:=\:2^{3R}\quad\Rightarrow\quad 2Q \:=\:3R \quad\Rightarrow\quad Q \:=\:\tfrac{3}{2}R \quad\Rightarrow\quad \log_4x \:=\:\tfrac{3}{2}\log_8x$

Substitute into [1]: .$\displaystyle 3\log_8x + \tfrac{3}{2}\log_8x + \log_8x \:=\:11 \quad\Rightarrow\quad \tfrac{11}{2}\log_8x \:=\:11$

Therefore: .$\displaystyle \log_8x \:=\:2\quad\Rightarrow\quad x \:=\:8^2 \quad\Rightarrow\quad\boxed{ x \:=\:64}$

• Jan 2nd 2010, 10:12 PM
Drexel28
Alternatively to Soroban. $\displaystyle \log_2(x)+\log_4(x)+\log_8(x)=\frac{\ln(x)}{\ln(2) }+\frac{\ln(x)}{\ln\left(2^2\right)}+\frac{\ln(x)} {\ln\left(2^3\right)}$$\displaystyle =\frac{6\ln(x)}{6\ln(2)}+\frac{3\ln(x)}{6\ln(x)}+\ frac{2\ln(x)}{6\ln(x)}=\frac{11\ln(x)}{6\ln(2)}$. Therefore our equation reduces to $\displaystyle \frac{11\ln(x)}{6\ln(2)}=11\implies \ln(x)=6\ln(2)=\ln\left(2^6\right)\implies x=2^6=64$
• Jan 5th 2010, 06:07 PM
ntrantrinh
Hi everyone, thanks for the solutions.
I'm still having some difficulties with the 1st and 2nd question.

Any guidance would be greatly appreciated :)
• Jan 6th 2010, 12:28 AM
Sudharaka
Dear ntrantrinh,

Please clarify about what problems you are having with the first and second parts. We will try to help.
• Jan 6th 2010, 09:31 AM
Soroban
Hello, ntrantrinh!

Here's #2 . . .

Quote:

If . $\displaystyle \begin{array}{ccc}a &= &\log_7 (11 - 6 \sqrt{2}) \\ b &=& \log_7 (45 + 29 \sqrt{2}) \end{array}$ . .find $\displaystyle 3a + 2b$ in simplest form.

We note that: .$\displaystyle \begin{array}{ccccccc}a & = &\log(11-6\sqrt{2}) &=& (3-\sqrt{2})^2 \\ b &=& \log_7(45 + 29\sqrt{2}) &=& (3+\sqrt{2})^3 \end{array}$

Then: .$\displaystyle 3a+2b \;=\;3\log_7(3-\sqrt{2})^2 + 2\log_7(3-\sqrt{2})^3$

. . . . . . . . . . $\displaystyle =\;6\log_7(3=\sqrt{2}) + 6\log_7(3+\sqrt{2})$

. . . . . . . . . . $\displaystyle =\;6\bigg[\log_7(3-\sqrt{2}) + \log_7(3+\sqrt{2})\bigg]$

. . . . . . . . . . $\displaystyle =\;6\,\log_7\!\bigg[(3-\sqrt{2})(3+\sqrt{2})\bigg]$

. . . . . . . . . . $\displaystyle =\;6\,\underbrace{\log_7(7)}_{\text{This is 1}}$

. . . . . . . . . . $\displaystyle =\; 6$

• Jan 10th 2010, 04:39 PM
ntrantrinh
Quote:

Originally Posted by Soroban
Hello, ntrantrinh!

Here's #2 . . .

We note that: .$\displaystyle \begin{array}{ccccccc}a & = &\log(11-6\sqrt{2}) &=& (3-\sqrt{2})^2 \\ b &=& \log_7(45 + 29\sqrt{2}) &=& (3+\sqrt{2})^3 \end{array}$

Then: .$\displaystyle 3a+2b \;=\;3\log_7(3-\sqrt{2})^2 + 2\log_7(3-\sqrt{2})^3$

. . . . . . . . . . $\displaystyle =\;6\log_7(3=\sqrt{2}) + 6\log_7(3+\sqrt{2})$

. . . . . . . . . . $\displaystyle =\;6\bigg[\log_7(3-\sqrt{2}) + \log_7(3+\sqrt{2})\bigg]$

. . . . . . . . . . $\displaystyle =\;6\,\log_7\!\bigg[(3-\sqrt{2})(3+\sqrt{2})\bigg]$

. . . . . . . . . . $\displaystyle =\;6\,\underbrace{\log_7(7)}_{\text{This is 1}}$

. . . . . . . . . . $\displaystyle =\; 6$

how did you get those logs to equal $\displaystyle (3-\sqrt{2})^2$ and $\displaystyle (3+\sqrt{2})^3$