Logarithmic Problem

Question

!!81^(1/(log_5\3)) + 27^(log_9\36) + 3^(4/log_7\9)!!

!!a)" "49!!

!!b)" "625!!

!!c)" "216!!

!!d)" "890!!

Answer

Lets try to simplify this equation so that we can easily calculate it.

!!81^(1/(log_5\3)) + 27^(log_9\36) + 3^(4/log_7\9)!!....... (i)

We know that !!log_a\b = 1/log_b\a!!

Using this simple concept in equation (i), we get

!!=(3^4)^(log_3\5) + (3^3)^(log_(3^2)(6^2)) + 3^(4log_9\7)!!

!!=(3)^(4log_3\5) + (3)^(3log_(3^2)(6^2)) + 3^(4log_(3^2)\7!!

Also we know that, !!log_(a^n)\b = (1/n)log_a\b!!

and !!log_a(b^n) = nlog_a\b!!

!!=(3)^(log_3(5^4)) + (3)^(3xx(2/2)log_3\6) + 3^(4xx(1/2)log_3\7!!

Since, !!a^(log_a\b) = b!!

Using this property, we get

!!=5^4 + (3)^(3log_3\6) + 3^(2log_3\7)!!

!!=5^4 + (3)^(log_3(6^3)) + 3^(log_3(7^2))!!

!!=5^4 + 6^3 + 7^2!!

!!= 625 + 216 + 49!!

!!=890!!

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