Question

Va rog mult rrpede dau coroană!!​

Answer 1

Sper sa înțelegi ceva coroană?

Answer 2

$\it a^{log_a b}=b\ \ \ \ \ (*)$

Vom transforma exponentul primului termen al sumei

$\it \dfrac{1}{log_5 9}=log_9 5=\dfrac{log_{81} 5}{log_{81} 9}=\dfrac{log_{81} 5}{log_{81} 81^{\frac{1}{2}}}=\dfrac{log_{81} 5}{\dfrac{1}{2}}=2log_{81} 5=log_{81}25$

Primul termen al sumei devine:

$\it 81^{log_{81}25}=25$

Vom transforma exponentul celui de al doilea termen al sumei

$\it log_9 16=\dfrac{log_3 16}{log_3 9}=\dfrac{log_3 16}{log_3 3^2}=\dfrac{log_3 16}{2}=\dfrac{1}{2}log_3 16=log_3 16^{\frac{1}{2}}=log_34$

Al doilea termen al sumei devine:

$\it 3^{log_3 4}=4$

Vom transforma exponentul celui de al treilea termen al sumei

$\it \dfrac{2}{log_7 9}=\dfrac{2}{\dfrac{log_3 9}{log_3 7}}=\dfrac{2}{\dfrac{2}{log_3 7}}=2:\dfrac{2}{log_3 7}=2\cdot\dfrac{log_3 7}{2}=log_3 7$

Al treilea termen al sumei devine:

$\it 3^{log_3 7}=7$

Suma va fi : 25 + 4 + 7 = 36