Question

Prove that 1/logx yz+1 + 1/logy zx+1 + 1/logz xy+1 = 1

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Answer 1

Step-by-step explanation:

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

=1/{logyz)/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)