# Math Help - Logarithms

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

2. #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:

$\log_{a}x^r=r\log_{a}x$ 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:

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

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:

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

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

$11\log_{2}x=66$

$\log_{2}x=6$

$2^6=x$

$x=64$

3. 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.

4. 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 $\sqrt[29]{2}$ sorry i'll correct that now.

5. Hello, ntrantrinh!

Here's my approach to #3 . . .

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

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

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

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

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

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

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

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

6. Alternatively to Soroban. $\log_2(x)+\log_4(x)+\log_8(x)=\frac{\ln(x)}{\ln(2) }+\frac{\ln(x)}{\ln\left(2^2\right)}+\frac{\ln(x)} {\ln\left(2^3\right)}$ $=\frac{6\ln(x)}{6\ln(2)}+\frac{3\ln(x)}{6\ln(x)}+\ frac{2\ln(x)}{6\ln(x)}=\frac{11\ln(x)}{6\ln(2)}$. Therefore our equation reduces to $\frac{11\ln(x)}{6\ln(2)}=11\implies \ln(x)=6\ln(2)=\ln\left(2^6\right)\implies x=2^6=64$

7. Hi everyone, thanks for the solutions.
I'm still having some difficulties with the 1st and 2nd question.

Any guidance would be greatly appreciated

8. Dear ntrantrinh,

Please clarify about what problems you are having with the first and second parts. We will try to help.

9. Hello, ntrantrinh!

Here's #2 . . .

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

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

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

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

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

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

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

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

10. Originally Posted by Soroban
Hello, ntrantrinh!

Here's #2 . . .

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

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

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

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

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

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

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

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