What is 1-log_{4}(m-1)+log_{4}(m+1) ?

The solution to 1-log_{4}(m-1)+log_{4}(m+1) is 1+log_{4}((m+1)/(m-1))

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

simplify 1-log_{4}(m-1)+log_{4}(m+1)

simplify

1 - lo g_{4} (m - 1) + lo g_{4} (m + 1)

1 + lo g_{4} (\frac{m + 1}{m - 1})

Solution steps

1 - lo g_{4} (m - 1) + lo g_{4} (m + 1)

Apply log rule:

lo g_{a} (x) - lo g_{a} (y) = lo g_{a} (\frac{x}{y})

lo g_{4} (m + 1) - lo g_{4} (m - 1) = lo g_{4} (\frac{m + 1}{m - 1})

= 1 + lo g_{4} (\frac{m + 1}{m - 1})

c lo g_{a} (x) + 2 lo g_{a} (y) - 4 lo g_{a} (x)

lo g_{10} (5 \frac{6}{x ^{7}})

lo g_{5} (\frac{25 b ^{5} y ^{2} - 1}{y ( x + 1 ) ^{2}})

ln (\frac{1}{e ^{8}}) + ln (8) e^{3}

ln (x^{6}) - 2 ln (x^{3})

What is 1-log_{4}(m-1)+log_{4}(m+1) ?
The solution to 1-log_{4}(m-1)+log_{4}(m+1) is 1+log_{4}((m+1)/(m-1))