8 log base8 to √2 * log base √8 to 8
Answers
Answer:
log 8 base 2+log 8 base 4+log 8base 16.
By change of bases, we can write log a base b as log a base x divided by log b base x.
log 8 base 2 can be written as log 2^3 base 2 divided by log 2 base 2 ...(*)
log 8 base 4 = log 2^3 base 2 divided by log 2^2 base 2 ...(**)
log 8 base 16 = log 2^3 base 2 divided by log 2^4 base 2 ...(***)
By the rule, log a ^m = m log a,
(*) can be written as 3log 2 base 2 divided by log 2 base 2 ..(****)
(**) can be written as 3log 2 base 2 divided by 2 log 2 base 2 ..(*****)
(***) can be written as 3log 2 base 2 divided by 4 log 2 base 2 ..(*****)
Taking 3 log 2 base 2 as common,
3 log 2 base 2{ (1/log 2 base 2 )+(1/2 log 2 base 2 )+(1/ 4 log 2 base 2 )}
we know that log a base a is equal to 1.
therefore, 3{ 1 + (1/2)+(1/4)}
on simplifying, we get,21/4
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What is log 7 in base 2?
8 can be expressed as 2^3 in terms of 2 whereas in terms of 4, 8 can be written as 4^(3/2) and 8 can written as 16^(3/4). (…..can provide an explaination how I got this, if you want)
Now using the property- log u^n to the base a = n times log u to the base a we can simplify the question as:
log 2^3 to the base 2 + log 4^(3/2) + log 16^(3/4)
= 3.log 2 to the base 2 + (3/2).log 4 to the base 4 + (3/4).log 16 to the base 16
Finally using the property- log a to the base a=1 we get:
= 3 + 3/2 + 3/4 = 21/4 as the final answer.
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Thans to the base changing formula:
log C base A = log C base B / log A base B
you get :
log 8 base 2 + [ log 8 base 2 / log 4 base 2] + [ log 8 base 2 / log 16 base 2] ;
Now you easily solve all the logs by their definition and get:
= 3 + [ 3 / 2 ] + [ 3 / 4] = (12 + 6 +3)/4 = 21 /4 !
That's all! If you have any doubts just write me! ;)
log2(8)+log4(8)+log16(8)
=3+ln(8)ln(4)+ln(8)ln(16)
=3+3ln(2)2ln(2)+3ln(2)4ln(2)
=3+32+34
=124+64+34
=214
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log8 to base2 + log8 to base 4 + log8 to base 16
= log8 to base2 + log8 to base 2^2 + log8 to base 2^4
= log2^3 to base 2 + log2^3 to base 2^2 + log2^3 to base 2^4
= using formula: log b^c to base a = c×log b to base a & log b to base a^m = ( log b to base a ) / m
= 3× log2 to base 2 + (3× log2 to base 2)/2 + (3× log2 to base 2)/4
= using formula : log a to base a = 1
= 3 + 3/2 + 3/4
= 21/4
Why do toppers choose BYJU'S for JEE preparation?
Let log8(base2)=x i.e2^x=8=>2^x=2^3=>x=3
log8(base4)=y=>4^y=8=>2^2y=2^3
=>2y=3=>y=3/2
log8(base16)=z=>16^z=8=>2^4z=2^3
=>4z=3=>z=3/4
x+y+z=3+3/2+3/4=21/4.
log 8 to the base 2 can be written as log 8/log 2
also log 8=3 log 2
log 4 = 2 log 2
log 16 = 4 log 2
therefore (log 8/log 2)+(log 8/ log 4) + (log 8/log 16)
=3+(3/2)+(3/4)=21/4
What is (log23)⋅(log34)⋅(log45)⋅(log56)⋅(log67)⋅(log78)?
What is log 8/log 8?
How is log 5 to the base 2?
How do I solve log base 16 of 1/2?
What is log 7 in base 2?
How do you calculate log base 2 (-1/4)?
What is the solution for - log 2 base 4 × log 4 base 6 × log 6 base 8 × log 8 base 10?
How can I solve this log base: 2 of x +log base 4 to x = 2?
How do I solve log 36 to the base 16 = 1 - log x to the base of 4?
Is log 7 base 2 rational?
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Step-by-step explanation: