#Can someone help me with these log issues?

83 messages · Page 1 of 1 (latest)

craggy topazBOT
strong tapir
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explain to a dumb guy like me

vocal ginkgo
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That looks kinda stressful

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What grade are you?

worn hazel
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First one

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The base is the cube root of 2

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Its 2^1/3

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So you can just remove the exponent

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1/1/3

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Thats 3

vocal ginkgo
worn hazel
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And we have the square root of 1/64

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64 is 2^6

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Sqrt((1/2)^6)

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So we get 3 . log 2 ((1/2)^6)

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We can remove the exponent (6)

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6.3 . Log 2 (1/2)

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18 . Log 2 (1/2)

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Log 2 (1/2) is the same as 2^x = 1/2

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x = -1

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so -18

vocal ginkgo
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Huh

worn hazel
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oh yeah

vocal ginkgo
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We are both wrong

worn hazel
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Yeah lol

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Ill do it on paper

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I got -44/5

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On first one

vocal ginkgo
vocal ginkgo
worn hazel
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I see

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happens

vocal ginkgo
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Got -44/5 as well

worn hazel
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K nice

vocal ginkgo
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Looks right

worn hazel
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@strong tapir u know how to solve these?

steady flicker
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Can you post the "entire" question? Part of it (at the top) is missing. Also, please translate the sentence in English.

strong tapir
strong tapir
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and need the step by step solution

worn hazel
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right

strong tapir
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sorry

worn hazel
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ill help with the rest of these

strong tapir
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I am Brazilian, education is bad here

strong tapir
worn hazel
strong tapir
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a

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KKKKKKKK

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que coincidência

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mn se vc me ajudar ia ser mt bom

worn hazel
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vou ajudar

strong tapir
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blz tmj

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preciso das 3 ate a 10

worn hazel
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vou conferir aqui

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é isso msm

strong tapir
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ok vlw

worn hazel
strong tapir
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?

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ou 3/4

worn hazel
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-3/4

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tinha o sinal de menos do log

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então fica -(3/4)

strong tapir
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mano obrigado devdd

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salvou mt mano tmj

worn hazel
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tmj

tall dust
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@strong tapir Qual era a questão mano?

strong tapir
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so resolver

tall dust