How do you condense #2log_3x-3log_3y+log_3 8#?

Question

Sadie Ackerman · Accepted Answer

#log_3((8x^2)/y^3)# Using the #color(blue)"laws of logarithms"# #color(orange)"Reminder " color(red)(|bar(ul(color(white)(a/a)color(black)(logx+logy=log(xy))color(white)(a/a)|)))" and "# #color(red)(|bar(ul(color(white)(a/a)color(black)(logx-logy=log(x/y))color(white)(a/a)|)))# #color(red)(|bar(ul(color(white)(a/a)color(black)(log x^nhArrnlogx)color(white)(a/a)|)))# #rArr2log_3 x=log_3 x^2" and " 3log_3 y=log_3 y^3# #rArrlog_3 x^2+log_3 8=log_3(8x^2)# and #log_3(8x^2)-log_3y^3=log_3((8x^2)/y^3)#

Lincoln Anderson · Answer

undefined To condense the expression $2\log_3 x - 3\log_3 y + \log_3 8$:

1. Apply the properties of logarithms, specifically the power rule and the product rule.
2. Use the power rule to bring the coefficients as exponents inside the logarithm.
3. Use the product rule to combine multiple logarithms with the same base into a single logarithm.
4. Simplify the resulting expression.

Following these steps:

$2\log_3 x - 3\log_3 y + \log_3 8$  
$= \log_3 x^2 - \log_3 y^3 + \log_3 8$  
$= \log_3 \left(\frac{x^2}{y^3}\right) + \log_3 8$  
$= \log_3 \left(\frac{x^2 \cdot 8}{y^3}\right)$  
$= \log_3 \left(\frac{8x^2}{y^3}\right)$

So, the condensed form of $2\log_3 x - 3\log_3 y + \log_3 8$ is $ \log_3 \left(\frac{8x^2}{y^3}\right) $.