# Thread: Please guide me along this logarithms sum(on expressing my answers in terms of p & q)

1. ## Please guide me along this logarithms sum(on expressing my answers in terms of p & q)

Hi ,
I got stuck doing this log question:
Given that log base 2 x=p and log base 4 y =q, express answer in terms of p and q.

a) log base x 4y
b)x^2y

Help is appreciated because I need to prepare for an upcoming test. Thanks

2. Originally Posted by AeroScizor
Hi ,
I got stuck doing this log question:
Given that log base 2 x=p and log base 4 y =q, express answer in terms of p and q.

a) log base x 4y
b)x^2y

Help is appreciated because I need to prepare for an upcoming test. Thanks
Hi AeroScizor,

Part (a)

$log_2x=p$

Now slide the "2" over to the "p" and drop the log

$x=_2p$

$x=2^p$

$log_4y=q\ \Rightarrow\ y=4^q\ \Rightarrow\ 4y=(4)4^q=4^{q+1}$

The change of base formula is $log_ax=\frac{log_bx}{log_ba}$

$log_x(4y)=\frac{log_2(4y)}{log_2(x)}$

$log_2(4y)=log_2\left(4^{q+1}\right)=z\ \Rightarrow\ 2^z=4^{q+1}=2^{2(q+1)}$

Therefore $log_2(4y)=z=2(q+1)$

Hence $log_x(4y)=\frac{2(q+1)}{p}$

Part (b)

$x^2y=\left(2^p\right)2^p\left(4^q\right)=2^{p+p}\l eft(2^2\right)^q=2^{2p}2^{2q}=2^{2p+2q}=2^{2(p+q)} =4^{p+q}$

If you meant $x^{2y}$

$2y=(2)4^q=(2)2^{2q}=2^{2q+1}$

$x^{2y}=\left(2^p\right)^{2^{2q+1}}$

Edit, sorry! I wrote the logs inverted earlier!!

3. Hello, AeroScizor!

I got a different answer for (a).

Given: . $\log_2(x)\:=\:p\:\text{ and }\:\log_4(y) \:=\:q$

(a) Express $\log_x(4y)$ in terms of $p\text{ and }q.$
Note that: . $\log_4(y) \:=\:q \quad\Rightarrow\quad 4^q \:=\:y \quad\Rightarrow\quad (2^2)^q \:=\:y \quad\Rightarrow\quad 2^{2q} \:=\:y$

. . Hence: . $\log_2(y) \:=\:2q$

Base-change Formula: . $\log_x(4y) \;=\;\frac{\log_2(4y)}{\log_2(x)} \quad\Rightarrow\quad \frac{\log_2(4) + \log_2(y)}{\log_2(x)} \;=\;\frac{2 + 2q}{p}$

4. Originally Posted by Soroban
Hello, AeroScizor!

I got a different answer for (a).

Note that: . $\log_4(y) \:=\:q \quad\Rightarrow\quad 4^q \:=\:y \quad\Rightarrow\quad (2^2)^q \:=\:y \quad\Rightarrow\quad 2^{2q} \:=\:y$

. . Hence: . $\log_2(y) \:=\:2q$

Base-change Formula: . $\log_x(4y) \;=\;\frac{\log_2(4y)}{\log_2(x)} \quad\Rightarrow\quad \frac{\log_2(4) + \log_2(y)}{\log_2(x)} \;=\;\frac{2 + 2q}{p}$

5. Originally Posted by Archie Meade
Hi AeroScizor,

Part (a)

$log_2x=p$

Now slide the "2" over to the "p" and drop the log

$x=_2p$

$x=2^p$

$log_4y=q\ \Rightarrow\ y=4^q\ \Rightarrow\ 4y=(4)4^q=4^{q+1}$

The change of base formula is $log_ax=\frac{log_bx}{log_ba}$

$log_x(4y)=\frac{log_2(4y)}{log_2(x)}$

$log_2(4y)=log_2\left(4^{q+1}\right)=z\ \Rightarrow\ 2^z=4^{q+1}=2^{2(q+1)}$

Therefore $log_2(4y)=z=2(q+1)$

Hence $log_x(4y)=\frac{2(q+1)}{p}$

Part (b)

$x^2y=\left(2^p\right)2^p\left(4^q\right)=2^{p+p}\l eft(2^2\right)^q=2^{2p}2^{2q}=2^{2p+2q}=2^{2(p+q)} =4^{p+q}$

If you meant $x^{2y}$

$2y=(2)4^q=(2)2^{2q}=2^{2q+1}$

$x^{2y}=\left(2^p\right)^{2^{2q+1}}$

Edit, sorry! I wrote the logs inverted earlier!!