$ log_2 6^0.5x - 1/4 = 8 $

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It is known that
$$ log_2 6^0.5x - 1/4 = 8 $$
What is $32x$? Put the answer as an integer.



Attempt :



$$ 2^8 = 6^0.5x - 1/4 implies 2^8 times 64 6^16 =6^32 x $$



But I cannot seem to write power of 2 in form of power of six. How to find $32x$? Thanks.




logarithms




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    The solution is not an integer. You can check that graphically by plotting $y=log_2 6^fracx2-frac14$.
    – csch2
    Jul 16 at 17:04















up vote
1
down vote

favorite












It is known that
$$ log_2 6^0.5x - 1/4 = 8 $$
What is $32x$? Put the answer as an integer.



Attempt :



$$ 2^8 = 6^0.5x - 1/4 implies 2^8 times 64 6^16 =6^32 x $$



But I cannot seem to write power of 2 in form of power of six. How to find $32x$? Thanks.




logarithms




share|cite|improve this question

















  • 1




    The solution is not an integer. You can check that graphically by plotting $y=log_2 6^fracx2-frac14$.
    – csch2
    Jul 16 at 17:04













up vote
1
down vote

favorite









up vote
1
down vote

favorite











It is known that
$$ log_2 6^0.5x - 1/4 = 8 $$
What is $32x$? Put the answer as an integer.



Attempt :



$$ 2^8 = 6^0.5x - 1/4 implies 2^8 times 64 6^16 =6^32 x $$



But I cannot seem to write power of 2 in form of power of six. How to find $32x$? Thanks.




logarithms




share|cite|improve this question













It is known that
$$ log_2 6^0.5x - 1/4 = 8 $$
What is $32x$? Put the answer as an integer.



Attempt :



$$ 2^8 = 6^0.5x - 1/4 implies 2^8 times 64 6^16 =6^32 x $$



But I cannot seem to write power of 2 in form of power of six. How to find $32x$? Thanks.





logarithms





share|cite|improve this question












share|cite|improve this question




share|cite|improve this question








edited Jul 16 at 16:48
























asked Jul 16 at 16:44









Arief

1,3311522




1,3311522







  • 1




    The solution is not an integer. You can check that graphically by plotting $y=log_2 6^fracx2-frac14$.
    – csch2
    Jul 16 at 17:04













  • 1




    The solution is not an integer. You can check that graphically by plotting $y=log_2 6^fracx2-frac14$.
    – csch2
    Jul 16 at 17:04








1




1




The solution is not an integer. You can check that graphically by plotting $y=log_2 6^fracx2-frac14$.
– csch2
Jul 16 at 17:04





The solution is not an integer. You can check that graphically by plotting $y=log_2 6^fracx2-frac14$.
– csch2
Jul 16 at 17:04











1 Answer
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In general, $log_a b^c = clog_a b$. Applying that:



$$log_2 6^0.5x-1/4 = left(dfracx2-dfrac14 right)log_2 6 = 8$$



Divide both sides by $log_2 6$, and $dfrac14$ to both sides, and multiply both sides by 2:



$$x = 2left(dfrac8log_2 6+dfrac14right)$$



$$32x = 64left( dfrac8log_2 6 + dfrac14right)$$






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  • Thanks. I edited the Q, the result must be in integer. (??)
    – Arief
    Jul 16 at 16:49










  • $$(log_6 8)/32=0.03626745068...$$ I don't know how you would get an integer for x. OOPS, the question has changed since I posted. Never mind.
    – poetasis
    Jul 16 at 17:06











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1 Answer
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1 Answer
1






active

oldest

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active

oldest

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active

oldest

votes








up vote
1
down vote



accepted










In general, $log_a b^c = clog_a b$. Applying that:



$$log_2 6^0.5x-1/4 = left(dfracx2-dfrac14 right)log_2 6 = 8$$



Divide both sides by $log_2 6$, and $dfrac14$ to both sides, and multiply both sides by 2:



$$x = 2left(dfrac8log_2 6+dfrac14right)$$



$$32x = 64left( dfrac8log_2 6 + dfrac14right)$$






share|cite|improve this answer





















  • Thanks. I edited the Q, the result must be in integer. (??)
    – Arief
    Jul 16 at 16:49










  • $$(log_6 8)/32=0.03626745068...$$ I don't know how you would get an integer for x. OOPS, the question has changed since I posted. Never mind.
    – poetasis
    Jul 16 at 17:06















up vote
1
down vote



accepted










In general, $log_a b^c = clog_a b$. Applying that:



$$log_2 6^0.5x-1/4 = left(dfracx2-dfrac14 right)log_2 6 = 8$$



Divide both sides by $log_2 6$, and $dfrac14$ to both sides, and multiply both sides by 2:



$$x = 2left(dfrac8log_2 6+dfrac14right)$$



$$32x = 64left( dfrac8log_2 6 + dfrac14right)$$






share|cite|improve this answer





















  • Thanks. I edited the Q, the result must be in integer. (??)
    – Arief
    Jul 16 at 16:49










  • $$(log_6 8)/32=0.03626745068...$$ I don't know how you would get an integer for x. OOPS, the question has changed since I posted. Never mind.
    – poetasis
    Jul 16 at 17:06













up vote
1
down vote



accepted







up vote
1
down vote



accepted






In general, $log_a b^c = clog_a b$. Applying that:



$$log_2 6^0.5x-1/4 = left(dfracx2-dfrac14 right)log_2 6 = 8$$



Divide both sides by $log_2 6$, and $dfrac14$ to both sides, and multiply both sides by 2:



$$x = 2left(dfrac8log_2 6+dfrac14right)$$



$$32x = 64left( dfrac8log_2 6 + dfrac14right)$$






share|cite|improve this answer













In general, $log_a b^c = clog_a b$. Applying that:



$$log_2 6^0.5x-1/4 = left(dfracx2-dfrac14 right)log_2 6 = 8$$



Divide both sides by $log_2 6$, and $dfrac14$ to both sides, and multiply both sides by 2:



$$x = 2left(dfrac8log_2 6+dfrac14right)$$



$$32x = 64left( dfrac8log_2 6 + dfrac14right)$$







share|cite|improve this answer













share|cite|improve this answer



share|cite|improve this answer











answered Jul 16 at 16:47









InterstellarProbe

2,217518




2,217518











  • Thanks. I edited the Q, the result must be in integer. (??)
    – Arief
    Jul 16 at 16:49










  • $$(log_6 8)/32=0.03626745068...$$ I don't know how you would get an integer for x. OOPS, the question has changed since I posted. Never mind.
    – poetasis
    Jul 16 at 17:06

















  • Thanks. I edited the Q, the result must be in integer. (??)
    – Arief
    Jul 16 at 16:49










  • $$(log_6 8)/32=0.03626745068...$$ I don't know how you would get an integer for x. OOPS, the question has changed since I posted. Never mind.
    – poetasis
    Jul 16 at 17:06
















Thanks. I edited the Q, the result must be in integer. (??)
– Arief
Jul 16 at 16:49




Thanks. I edited the Q, the result must be in integer. (??)
– Arief
Jul 16 at 16:49












$$(log_6 8)/32=0.03626745068...$$ I don't know how you would get an integer for x. OOPS, the question has changed since I posted. Never mind.
– poetasis
Jul 16 at 17:06





$$(log_6 8)/32=0.03626745068...$$ I don't know how you would get an integer for x. OOPS, the question has changed since I posted. Never mind.
– poetasis
Jul 16 at 17:06













 

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