What is ln((1/2+x/4)^a*(1/4-x/4)^{b+c}*(x/4)^d) ?

The solution to ln((1/2+x/4)^a*(1/4-x/4)^{b+c}*(x/4)^d) is aln(1/2+x/4)+(b+c)ln(1/4-x/4)+dln(x/4)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

simplify ln((1/2+x/4)^a(1/4-x/4)^{b+c}(x/4)^d)

simplify

ln ((\frac{1}{2} + \frac{x}{4})^{a} \cdot (\frac{1}{4} - \frac{x}{4})^{b + c} \cdot (\frac{x}{4})^{d})

a ln (\frac{1}{2} + \frac{x}{4}) + (b + c) ln (\frac{1}{4} - \frac{x}{4}) + d ln (\frac{x}{4})

Solution steps

ln ((\frac{1}{2} + \frac{x}{4})^{a} (\frac{1}{4} - \frac{x}{4})^{b + c} (\frac{x}{4})^{d})

Apply log rule:

lo g_{c} (ab) = lo g_{c} (a) + lo g_{c} (b)

ln ((\frac{1}{2} + \frac{x}{4})^{a} (\frac{1}{4} - \frac{x}{4})^{b + c} (\frac{x}{4})^{d}) = ln ((\frac{1}{2} + \frac{x}{4})^{a}) + ln ((\frac{1}{4} - \frac{x}{4})^{b + c}) + ln ((\frac{x}{4})^{d})

= ln ((\frac{x}{4} + \frac{1}{2})^{a}) + ln ((- \frac{x}{4} + \frac{1}{4})^{b + c}) + ln ((\frac{x}{4})^{d})

Apply log rule:

lo g_{a} (x^{b}) = b \cdot lo g_{a} (x)

ln ((\frac{1}{2} + \frac{x}{4})^{a}) = a ln (\frac{1}{2} + \frac{x}{4})

= a ln (\frac{1}{2} + \frac{x}{4}) + ln ((\frac{1}{4} - \frac{x}{4})^{b + c}) + ln ((\frac{x}{4})^{d})

Apply log rule:

lo g_{a} (x^{b}) = b \cdot lo g_{a} (x)

ln ((\frac{1}{4} - \frac{x}{4})^{b + c}) = (b + c) ln (\frac{1}{4} - \frac{x}{4})

= a ln (\frac{1}{2} + \frac{x}{4}) + (b + c) ln (\frac{1}{4} - \frac{x}{4}) + ln ((\frac{x}{4})^{d})

Apply log rule:

lo g_{a} (x^{b}) = b \cdot lo g_{a} (x)

ln ((\frac{x}{4})^{d}) = d ln (\frac{x}{4})

= a ln (\frac{1}{2} + \frac{x}{4}) + (b + c) ln (\frac{1}{4} - \frac{x}{4}) + d ln (\frac{x}{4})

\frac{3}{4} lo g_{8} (18)

\frac{ln ( 5 ) + ln ( 5 ^{2} ) + ln ( 3 ^{3} )}{ln ( 5 )}

ln (x^{3}) - ln (x^{2} + x)

\frac{13}{8} \cdot ln (\frac{13}{8}) - \frac{1}{\frac{13}{8}}

ln (2 \cdot 0.252)

What is ln((1/2+x/4)^a*(1/4-x/4)^{b+c}*(x/4)^d) ?
The solution to ln((1/2+x/4)^a*(1/4-x/4)^{b+c}*(x/4)^d) is aln(1/2+x/4)+(b+c)ln(1/4-x/4)+dln(x/4)