Simplify $2^{3 + 5 {log}_{2} x}$ $2^(3+5log_2x)$ ?

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Answer 1 · 2018-02-08T14:51:39

$2^{3 + 5 {log}_{2} x} = 8 x^{5}$ $2^(3+5log_2x)=8x^5$

Let $a^{{log}_{a} x} = u$ $a^(log_ax)=u$ . Then taking logarithm to the base $a$ $a$ on both sides we get

${log}_{a} x \times {log}_{a} a = {log}_{a} u$ $log_ax xxlog_aa=log_au$

or ${log}_{a} u = {log}_{a} x$ $log_au=log_ax$ and therefore $u = x$ $u=x$ i.e.

$a^{{log}_{a} x} = x$ $a^(log_ax)=x$

Using this in $2^{3 + 5 {log}_{2} x}$ $2^(3+5log_2x)$

= $2^{3} \times 2^{5 {log}_{2} x}$ $2^3xx2^(5log_2x)$

= $2^{3} \times {(2^{{log}_{2} x})}^{5}$ $2^3xx(2^(log_2x))^5$ - as $a^{m n} = {(a^{m})}^{n}$ $a^(mn)=(a^m)^n$ or ${(a^{n})}^{m}$ $(a^n)^m$

= $8 x^{5}$ $8x^5$

Simplify 23+5log2x2^(3+5log_2x)?