Question

Given that logx =m+n and logy=m-n,value of log10x/y^2 is expressed as

Answer 1

Answer:

Step-by-step explanation:

logx = m+n

logy = m-n

log(10x/y²) = log10x - logy²

                  = log10 + logx - 2logy

                  = 1 + m+n - 2(m-n)

                 = 1-m+3n

Answer 2

Answer:

-m+3n+1

Step-by-step explanation:

$log_{10} x = m + n$

$log_{10} y = m - n$

$log \frac{10x}{y^{2} }$ = $log_{10} 10x$$- log_{10} y^{2}$

$log \frac{10x}{y^{2} }$ = $log_{10} 10 + log_{10} x - log_{10}{y^{2}$

$log\frac{10x}{y^{2} }$ = $log_{10} 10 + log_{10} x - 2 log_{10} y$

$log\frac{10x}{y^{2}} = 1 + (m+n) - 2(m-n)$

$log \frac{10x}{y^{2}} = 1 + m +n - 2m + 2n$

$Hence Solved$