Math, asked by mithun1605, 3 months ago

prove that log a+log a²+log a³+...log a⁸=36 log a​

Answers

Answered by abdurrafay102003
0

Step-by-step explanation:

 log(a)  +  log( {a}^{2} )  +  log( {a}^{3} )  +  log( {a}^{8} )  = 36 log(a) \\

Using product law of logarithms,

 log(a \times  {a}^{2}  \times  {a}^{3}  \times  {a}^{8} )  = 36 log(a)

Using product law of indices,

 log( {a}^{1 + 2 + 3 + 8} )  = 36 log(a)  \\  log( {a}^{14} )  = 36 log(a)  \\ 14 log(a)  = 36 log(a)

I think your question is missing, that's why it couldn't be proved, but I tried. Please mark as brainliest will be highly appreciated :)

Answered by bson
0

Step-by-step explanation:

log a*a²*a³*...a⁸

= log a^(1+2+3+4+5+6+7+8)

=log a^(8×9/2)= log a³⁶ = 36 loga

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