SOLUTION: 4log{{{2}}}x + log{{{2}}}5 = log{{{2}}}405 What is the value of X in this logarithm? When I tried to work it I simplified the equation to log{{{2}}}x^4 + log{{{2}}}5 = log{{{2}}}40
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-> SOLUTION: 4log{{{2}}}x + log{{{2}}}5 = log{{{2}}}405 What is the value of X in this logarithm? When I tried to work it I simplified the equation to log{{{2}}}x^4 + log{{{2}}}5 = log{{{2}}}40
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Question 20542: 4logx + log5 = log405 What is the value of X in this logarithm? When I tried to work it I simplified the equation to logx^4 + log5 = log405. Then I subtracted x from both sides and was left with logx^4 = 400. Please walk me through this because even if I am doing it correctly I don't know what to do next. Answer by AnlytcPhil(1806) (Show Source):
You can put this solution on YOUR website! 4log2x + log25 = log2405 What is the value of x in this logarithm? When I tried
to work it I simplified the equation to log2x4 + log25 = log2405. Then I
subtracted x from both sides and was left with = 400. Please walk me through
this because even if I am doing it correctly I don't know what to do next.
`
One error was in thinking log2405 - log25 = log2(405-5) = log2(400). This is
wrong. The rule is
`
logBU - logBV = logB(U/V)
`
[NOT logB(U-V)]
`
So you should have gotten log2405 - log25 = log2(405/5) = log2(81).
`
This would have simplified to
`
x4 = 81
`
which you could have solved and gotten x = ±3 and then discarded the -3 since
logs cannot be taken of negative numbers.
`
However. also always try, if possible, to avoid using the rule N·logBX = logBXN
whenever it causes variables to be raised to higher powers. You can often
avoid this by dividing through by the value of N.
`
4log2x + log25 = log2405
`
Subtract log25 from both sides
`
4log2x = log2405 - log25
`
Use the rule logU - logBV = logB(U/V) on the right
`
4log2x = log2(405/5)
`
4log2x = log2(81)
`
Here is where you want to avoid using the rule N·logBX = logBXN, to avoid
causing the variable x to be raised to the fourth power. Let's work on the right
side some more instead, noting that 81 = 34.
`
4log2x = log2(34)
`
Now we can use logBXN = N·logBX on the right sides
`
4log2x = 4log23
`
Now we can divide both sides by 4.
`
log2x = log23
`
Raise both sides to the 2nd power, which is the same as dropping the log2's,
and we get
`
x = 3
`
Edwin
AnlytcPhil@aol.com
