Question

What is 4log1/2^w(2log1/2^u-3log1/2^v written as a single logarithm?

Answer 1

Given:

4log1/2^w (2log1/2^u-3log1/2^v)

Req'd:

Single logarithm = ?

Sol'n:

First remove the parenthesis,

4 log 1/2 (w) + 2 log 1/2 (u) - 3 log 1/2 (v)

Simplify each term,

Simplify the 4 log 1/2 (w) by moving the constant 4 inside the logarithm;
Simplify the 2 log 1/2 (u) by moving the constant 2 inside the logarithm;
Simplify the -3 log 1/2 (v) by moving the constant -3 inside the logarithm:

 log 1/2 (w^4)  + 2 log 1/2 (u) - 3 log 1/2 (v) 
log 1/2 (w^4) + log 1/2 (u^2) - log 1/2 (v^3)

We have to use the product property of logarithms which is log of b (x) + log of b (y) = log of b (xy):

Thus,

Log of 1/2 (w^4 u^2) - log of 1/2 (v^3) 

then use the quotient property of logarithms which is log of b (x)  - log of b (y) = log of b (x/y)

Therefore, 

log of 1/2 (w^4 u^2 / v^3)

and for the final step and answer, reorder or rearrange w^4 and u^2:

log of 1/2 (u^2 w^4 / v^3)  

Answer 2

Answer:

$4\log_{\frac{1}{2}}w+(2\log_{\frac{1}{2}}u-3\log_{\frac{1}{2}}v)=\log_{\frac{1}{2}}(\frac{w^4u^2}{v^3})$

Step-by-step explanation:

Given : Expression  $4\log_{\frac{1}{2}}w+(2\log_{\frac{1}{2}}u-3\log_{\frac{1}{2}}v)$

To write : As a single logarithm?

Solution :

$4\log_{\frac{1}{2}}w+(2\log_{\frac{1}{2}}u-3\log_{\frac{1}{2}}v)$  

Remove parenthesis,

$=4\log_{\frac{1}{2}}w+2\log_{\frac{1}{2}}u-3\log_{\frac{1}{2}}v$  

Simplify each term by applying logarithmic property, $a\log x=\log x^a$

$=\log_{\frac{1}{2}}w^4+\log_{\frac{1}{2}}u^2-\log_{\frac{1}{2}}v^3$  

Use the product property of logarithms, $\log_bx+\log_b y=\log_b (xy)$

$=\log_{\frac{1}{2}}w^4u^2-\log_{\frac{1}{2}}v^3$  

Use the quotient property of logarithms, $\log_bx-\log_b y=\log_b (\frac{x}{y})$

$=\log_{\frac{1}{2}}(\frac{w^4u^2}{v^3})$  

Therefore,

$4\log_{\frac{1}{2}}w+(2\log_{\frac{1}{2}}u-3\log_{\frac{1}{2}}v)=\log_{\frac{1}{2}}(\frac{w^4u^2}{v^3})$