1. ## Logarithms

Given that $x=log_2a$and $y=log_2b$, express the following in terms of $y$ and $x$

(ii) $log_{2a}\sqrt{ab^3}$

My attempt:

$x=log_2a$
$a=2^x$

$y=log_2b$
$b=2^y$

$log_{2a}\sqrt{ab^3}$= $log_{2(2^x)}\sqrt{2^x(2^{3y})}$

= $log_{2^{2x+1}}\sqrt{2^x}+log_{2^{2x+1}}\sqrt{2^{3y }}$

2. Why do you make it so complicated ?

$x = \log_2{(a)} \Rightarrow a = 2^x$
$y = \log_2{(b)} \Rightarrow b = 2^y$

$\log_2{\left ( a \sqrt{ab^3} \right )}$

$= \log_2{\left (2^x \sqrt{2^x (2^y)^3} \right )}$

$= \log_2{\left (2^x \sqrt{2^x 2^{3y}} \right )}$

$= \log_2{\left (2^x \sqrt{2^{x + 3y}} \right )}$

$= \log_2{\left (2^x \times 2^{\frac{x + 3y}{2}} \right )}$

$= \log_2{\left (2^{x + \frac{x + 3y}{2}} \right )}$

$= x + \frac{x + 3y}{2}$

$= \frac{2x}{2} + \frac{x + 3y}{2}$

$= \frac{2x + x + 3y}{2}$

$= \frac{3x + 3y}{2}$

$= \frac{3(x + y)}{2}$

$= \frac{3}{2}(x + y)$

3. $x=log_2a$
$a=2^x$

$y=log_2b$
$b=2^y$
It is excellent that you have memorized this transformation. It seems, however, that you have not learned how to use the information in the section.

Originally Posted by Punch
Given that $x=log_2a$and $y=log_2b$, express the following in terms of $y$ and $x$

(ii) $log_2a\sqrt{ab^3}$

My attempt:

$log_2a\sqrt{ab^3}$= $log_{2(2^x)}\sqrt{2^x(2^{3y})}$

1) You should not have done ANYTHING with the base of the logartihm. All the logarithms are Base 2.

2) There is no such thing as a superscript on a logarithm. Perhaps it was just a codign error.
= $log_2^{2x+1}\sqrt{2^x}+log_2^{2x+1}\sqrt{2^{3y}}$
3) How about if we just use rules of logarithms to solve this.

$log_{2}(a\sqrt{ab^3})$

$log_{2}((a)\cdot(\sqrt{ab^3}))$

$log_{2}(a) + log_{2}(\sqrt{ab^3})$

$log_{2}(a) + \frac{1}{2}log_{2}(ab^3)$

$log_{2}(a) + \frac{1}{2}(log_{2}(a) + log_{2}(b^3))$

$log_{2}(a) + \frac{1}{2}(log_{2}(a) + 3\cdot log_{2}(b))$

$log_{2}(a) + \frac{1}{2} \cdot log_{2}(a) + \frac{3}{2}\cdot log_{2}(b)$

You can finish?

4. You need not convert in to index form. Using the laws of logarithm you can simplify.

***********

5. Sorry but there was a typo error on the question i typed...

it should be

Given that $x=log_2a$ and $y=log_2b$, express the following in terms of $y$ and $x$

(ii) $log_{2a}\sqrt{ab^3}$

6. $\log_{2a}{\left ( \sqrt{ab^3} \right )}$

$= \log_{2^{x + 1}}{\left ( 2^{\frac{x + 3y}{2}} \right )}$

$= \log_{2^{x + 1}}{\left ( 2^{\frac{x + 3y}{2}} \right )}$

$= \frac{x + 3y}{2} \times \log_{2^{x + 1}}{\left ( 2 \right )}$

$= \frac{x + 3y}{2} \times \frac{\log_2{(2)}}{\log_2 {\left ( 2^{x + 1}\right ) }}$

$= \frac{x + 3y}{2} \times \frac{1}{x + 1}$

$= \frac{x + 3y}{2(x + 1)}$

$= \frac{1}{2} \times \frac{x + 3y}{x + 1}$

Does it make sense ?

7. Hello, Punch!

Given that: . $\begin{Bmatrix}
x&=& \log_2a \\ y &=& \log_2b \\ \end{Bmatrix}$
.[1]

express the following in terms of $x\text{ and }y:$

. . $(ii)\;\;\log_{2a}\sqrt{ab^3}$

Let: . $\log_{2a}\sqrt{ab^3} \:=\:P$ .[2]

Then: . $(2a)^P \;=\;\left(ab^3\right)^{\frac{1}{2}}$

Take logs (base 2): . $\log_2\left(2a\right)^P \;=\;\log_2\left(ab^3\right)^{\frac{1}{2}}$

. . . . . . . . . . . . . . $P\cdot\log_2(2a) \;=\;\frac{1}{2}\log_2\left(ab^3\right)$

. . . . . . . . $P\cdot\bigg[\log_2(2) + \log_2a\bigg] \;=\;\frac{1}{2}\bigg[\log_2a + \log_2\left(b^3\right)\bigg]$

. . . . . . . . . . . $P\cdot\bigg[1 + \log_2a\bigg] \;=\;\frac{1}{2}\bigg[\log_2a + 3\log_2b\bigg]$

Multiply by 2: . $P\cdot2\bigg[1 + \log_2a\bigg] \;=\;\log_2a + 3\log_2b$

. . . . . . . . . . . . . . . . . . . $P \;=\;\frac{\log_2a + 3\log_2b}{2(1+\log_2a)}$ .[3]

Equate [2] and [3]: . $\log_{2a}\sqrt{ab^3} \;=\;\frac{\log_2a + 3\log_2b}{2(1 + \log_2a)}$

. . . Substitute [1]: . $\log_{2a}\sqrt{ab^3} \;=\;\frac{x+3y}{2(1+x)}$

8. Originally Posted by Punch
Sorry but there was a typo error on the question i typed...

it should be

Given that $x=log_2a$ and $y=log_2b$, express the following in terms of $y$ and $x$

(ii) $log_{2a}\sqrt{ab^3}$
Using the change of base, rewrite the above problem.
[1/2log(ab^3) to the basr 2]/[log(2a) to the base2]
Now use the laws of logarithm to simplify.