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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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 log<sub>2</sub>405 - log<sub>2</sub>5 = log<sub>2</sub>(405-5) = log<sub>2</sub>(400). This is
wrong. The rule is
`
log_BU - log_BV = log_B(U/V)
`
[NOT log<sub>B</sub>(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·log_BX = log_BX^N
whenever it causes variables to be raised to higher powers. You can often
avoid this by dividing through by the value of N.
`
4log<sub>2</sub>x + log<sub>2</sub>5 = log<sub>2</sub>405
`
Subtract log₂5 from both sides
`
4log<sub>2</sub>x = log<sub>2</sub>405 - log<sub>2</sub>5
`
Use the rule logU - log_BV = log_B(U/V) on the right
`
4log<sub>2</sub>x = log<sub>2</sub>(405/5)
`
4log₂x = log₂(81)
`
Here is where you want to avoid using the rule N·log<sub>B</sub>X = log<sub>B</sub>X<sup>N</sup>, 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 = 3<sup>4</sup>.
`
4log₂x = log₂(3⁴)
`
Now we can use log<sub>B</sub>X<sup>N</sup> = N·log<sub>B</sub>X 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 log<sub>2</sub>'s,
and we get
`
x = 3
`
Edwin
AnlytcPhil@aol.com