if a=2 and b^2-ab=1 then wat is the value of log(a+b)(a^3+b^3)
Answers
Answer:
Since the logarithm of a fraction M/N is equal to the logarithm of the numerator diminished by the logarithm of the denominator, that is
log(M/N) = log M - log N ,
the given equation
log (a+b)/3 - log(a/2) - log(b/2) = log[(a+b)/3/(a/2)] - log(b/2)
=log[2(a+b)/3a] - log(b/2) = log[2(a+b)/3a/b/2]
= log [4(a+b)/3ab] ………………………………………………………(1)
Given a² + b² = 7ab,
Or, a² + b² + 2ab = 2ab + 7ab =9ab Or, (a+b)² = 9ab
Or, 3ab = (a+b)²/3 Substituting for 3ab in (1),
log (a+b)/3 - log(a/2) - log(b/2)
= log[4(a+b)/(a+b)²/3]
=log [12/(a+b)] (Answer)
If a² + b² = 7ab, prove that log [⅓(a+b)] = ½ [log a + log b] ?
How can we prove that if a2+b2=7ab and a>0,b>0 then loga+b3=12(loga+logb) ?
If a square +b square = 7ab, how do you prove that 1/2 (loga +logb) =log (a+b/3)?
If log 2 =a, log 3=b, log 7=c and 6^x=7^x+4, then what is the value of x?
If a^3 + b^3 = 0, then what is the value of log(a+b)-1/2 (loga+logb+log3)?
a^2+b^2 = 7ab => a^2+b^2+2ab = 9ab
=> (a+b)^2 = 9ab => a+b = 3√(ab)….. (1)
log{(a+b)/3}-log(a/2) -log(b/2)
= log {(a+b)/3}/{(a/2)(b/2)
= log 4(a+b)/3ab = log {4×3√(ab)/3ab}
= log 4/√(ab)
I hope this helps.
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a^2 + b^2 +2ab = 9ab
(a+b)^2 = √9ab
(a+b) = 3√ab
(a+b)/3= √ab
log(a+b)/3 = log√ab
log(a+b)/3 = log(ab)^1/2
log(a+b)/3 = 1/2[loga + logb]…
as log(ab) = log a+ log b
so the value is 0
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If a² + b² = 7ab, prove that log [⅓(a+b)] = ½ [log a + log b] ?
How can we prove that if a2+b2=7ab and a>0,b>0 then loga+b3=12(loga+logb) ?
If a square +b square = 7ab, how do you prove that 1/2 (loga +logb) =log (a+b/3)?
If log 2 =a, log 3=b, log 7=c and 6^x=7^x+4, then what is the value of x?
If a^3 + b^3 = 0, then what is the value of log(a+b)-1/2 (loga+logb+log3)?
If log {a+b} /2 = 1/2 (log a+ log b) , prove that a = b?
If log 2=0.3010, what is the value of log 500?
If log 3 base 2 = x , what is log 24 base 12?
How do I solve log base 3*log base 2 *log base root 5 (5^4)?
If log r^6=m and log r^3=n then what is log (r^r/2) equal to?
What is the value of log 2?
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Answer:
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