What is 1/5 ln((x^2-25)^5)+ln(e)-ln(1)-ln(x+5) ?

The solution to 1/5 ln((x^2-25)^5)+ln(e)-ln(1)-ln(x+5) is ln(x-5)+1

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simplify 1/5 ln((x^2-25)^5)+ln(e)-ln(1)-ln(x+5)

simplify

\frac{1}{5} ln ((x^{2} - 25)^{5}) + ln (e) - ln (1) - ln (x + 5)

ln (x - 5) + 1

Solution steps

\frac{1}{5} ln ((x^{2} - 25)^{5}) + ln (e) - ln (1) - ln (x + 5)

Apply log rule:

a lo g_{c} (b) = lo g_{c} (b^{a})

\frac{1}{5} ln ((x^{2} - 25)^{5}) = ln (((x^{2} - 25)^{5})^{\frac{1}{5}})

= ln (((x^{2} - 25)^{5})^{\frac{1}{5}}) + ln (e) - ln (1) - ln (x + 5)

Apply log rule:

lo g_{c} (a) - lo g_{c} (b) = lo g_{c} (\frac{a}{b})

ln (((x^{2} - 25)^{5})^{\frac{1}{5}}) - ln (x + 5) = ln \frac{( ( x ^{2} - 25 ) ^{5} ) ^{\frac{1}{5}}}{x + 5}

= ln \frac{( ( x ^{2} - 25 ) ^{5} ) ^{\frac{1}{5}}}{x + 5} + ln (e) - ln (1)

Simplify

\frac{( ( x ^{2} - 25 ) ^{5} ) ^{\frac{1}{5}}}{x + 5} : x - 5

= ln (x - 5) + ln (e) - ln (1)

Apply log rule:

lo g_{a} (a) = 1

= ln (x - 5) + 1 - ln (1)

Apply log rule:

lo g_{a} (1) = 0

= ln (x - 5) + 1 - 0

Add/Subtract the numbers:

1 - 0 = 1

= ln (x - 5) + 1

ln (R) - ln (\frac{1}{k})

4.7447 + lo g_{10} (\frac{0.5}{0.75})

ln (\frac{10}{12.5}) + ln (\frac{8}{10})

ln (\frac{x}{x - 2})

lo g_{2} (\frac{16}{x ^{4}})

What is 1/5 ln((x^2-25)^5)+ln(e)-ln(1)-ln(x+5) ?
The solution to 1/5 ln((x^2-25)^5)+ln(e)-ln(1)-ln(x+5) is ln(x-5)+1