Write the expression as a single logarithm.
4log6x-1/5 log6z + 2log6y
4log6(x−1)5log6z+2log6y\frac{4log_{6}(x-1)}{5log_{6}z+2log_{6}y}5log6z+2log6y4log6(x−1)
=>log6(x−1)4log6z5+log6y2=>\frac{log_{6}(x-1)⁴}{log_{6}z⁵+log_{6}y²}=>log6z5+log6y2log6(x−1)4
=>log6(x−1)4log6z5y2=>\frac{log_{6}(x-1)⁴}{log_{6}z⁵y²}=>log6z5y2log6(x−1)4
Recall that logxalogxb=logba\frac{log_{x} a}{log_{x}b} = log_{b}alogxblogxa=logba
=>log6(x−1)4log6z5y2=logz5y(x−1)4=>\frac{log_{6}(x-1)⁴}{log_{6}z⁵y²}={log_{z⁵y}(x-1)⁴}=>log6z5y2log6(x−1)4=logz5y(x−1)4
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