Question

here's a question for fun:

$log_{2}x- log_{4}x- log_{16}x- log_{64}x=1$

solve and show ALL work
no credit for guessing (guessers or people who make a bogus explanation will be reported)

Answer 1

X is Equal to 4096. seeing how you have all logs bases and no log = to, you must combine them. using your Log rules. then after that you will convert it to exponential form then just put it in a calculator to get the answer. 

This was the way i learned it in 10th grade. Im in 11th  now. Hope it helps!!

Answer 2

log2 x- log4 x-log16 x-log64 x=1

loga b=logc a / logc b

Therefore:

log2 x- (log2 x / log2 4) - (log2 x / log2 16) - (log2 x / log2 64)=log2 2

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loga n=x    ⇔a^x=n

log2 4=2      ⇔ 2²=4
log2 16=4    ⇔2⁴=16
log2 64=6    ⇔2⁶=64

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log2 x - (log2 x / 2)- (log2 x / 4)- log2 x / 6)=log2 2

least common multiple=12
12 log2 x-4log2 x - 3log2 x - log2 x=12log2 2

l------------------------------------------------------------------------------------

nloga x=loga x^n

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log2 x¹² - log2 x ⁴ - log2 x³ - log2 x= log2 2¹²

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loga n -loga p=loga n/p

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log2(x¹² / x⁴)-log2 (x³/x)=log2 2¹²
log2 x³ - log2 x²=log2 2¹²
log2 (x³ / x²) =log2 2¹²
log2 x=log2 2¹²

therefore: x=2¹²

Answer: x=2¹²

we check it out the answer.
log2 2¹² - log4 2¹² - log16 2¹²- log64 2¹²=12-6-3-2=1

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log2 2¹²=12  ⇔2¹²=2¹²
log4 2¹²=6    ⇔ 4⁶=2¹²
log16 2¹²=3  ⇔ 16³=12¹²
log64 2¹²=2 ⇔  64²=2¹²
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