If $k^{\log_{2}5}=16$ then find $k^{{(\log_{2}5})^2} = $?

Question

$$k^{\log_{2}5}=16$$ then find $$k^{{(\log_{2}5})^2}$$

Note: the exponent(entire log) is squared unlike the value Inside the log squared.

Answer 1

Let $x=k^{(\log_25)^2}$. Taking $\log_5$ on both sides,

$$\log_5 x=\log_5\left[k^{(\log_25)(\log_25)}\right]$$ $$\log_5 x=\log_25\cdot\log_5\left[k^{\log_25}\right]$$ $$\log_5 x=\log_25\cdot\log_516$$

$$\log_5x=\log_216$$ $$\log_5x=4$$ So$$x=5^4=625$$

Answer 2

Hint

Use exponent laws:

$$k^{a^2}=k^{a \cdot a}= \left(\color{blue}{k^a}\right)^a$$

In your case $a=\log_25$ and you know the value of $\color{blue}{k^a}$.

Or, put differently:

$$k^{\log_{2}5}=16$$

Raise both sides of this equality to the power $\log_25$: the left-hand side is then exactly what you need to find, the right-hand side is the answer (although you can simplify!).

Answer 3

Since $k=16^{\frac{1}{\log_25}},$ we obtain: $$k^{\log^2_25}=16^{\log_25}=2^{4\log_25}=2^{\log_2625}=625.$$

Answer 4

Let $\log_25=a,\implies2^a=5$

and $k^a=16$

$k^{a^2}=(k^a)^a=16^a=(2^a)^4=?$