Question

40) Find the
value of (1÷ logX(yz)+1 )+(1÷logY(zx)+1)+(1÷logZ(xy)+1)​

Answer 1

Solution:

Now, 1/log_x(yz) + 1/log_y(zx) + 1/log_z(xy)

= 1/{log(yz)/logx + 1} + 1/{log(zx)/logy + 1} + 1/{log(xy)/logz + 1}

= 1/{(logyz + logx)/logy} + 1/{(logzx + logy)/logy} + 1/{(logxy + logz)/logz}

= 1/{log(xyz) / logy} + 1/{log(xyz) / logy} + 1/{log(xyz) / logz}

= logy/log(xyz) + logy/log(xyz) + logz/log(xyz)

= (logy + logx + logz)/log(xyz)

= log(xyz) / log(xyz)

= 1

Rule:

1. log_a(b) = logb / loga

2. loga + logb + logc = log(abc)