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If $log_a {y^5} + 4 log_a {y^3} = 34$ , then a=?

\sqrt{y}

\frac{1}{y}

y^2

y

1 solution

Abhisek Mohanty
Apr 1, 2015

First let us simplify the LHS

$log_{a}y^{5}+ 4log_{a}y^{3}$

$=log_{a}y^{5} + log_{a}y^{3\times4}$

$=log_{a}y^{5} + log_{a}y^{12}$

$=log_{a}(y^{5})(y^{12})$

$=log_{a}y^{17}$

Now as LHS=RHS

$log_{a}y^{17} = 34$

$or, a^{34} = y^{17}$

$or, a^{2} = y$

$or, a = \sqrt{y}$

y must be greater than zero and different from 1

Константиновић Марко - 2 years, 7 months ago