Log question (1 Viewer)

Green Yoda · Jul 2, 2015

Need help with Question 3H!

leehuan · Jul 2, 2015

$\log _{ 10 }{ \sqrt { 40 } } \\ =\frac { 1 }{ 2 } \log _{ 10 }{ 40 } \\ =\frac { 1 }{ 2 } \left( \log _{ 10 }{ 4 } +\log _{ 10 }{ 10 } \right)$

Log laws used:
Coefficient<->Power
Sum<->Product

The next step is just to recognise that the log base n of n is just 1

Note that a square root just means to take a number to the power of a half.

Green Yoda · Jul 2, 2015

leehuan said:
$\log _{ 10 }{ \sqrt { 40 } } \\ =\frac { 1 }{ 2 } \log _{ 10 }{ 40 } \\ =\frac { 1 }{ 2 } \left( \log _{ 10 }{ 4 } +\log _{ 10 }{ 10 } \right)$

Log laws used:
Coefficient<->Power
Sum<->Product

The next step is just to recognise that the log base n of n is just 1

Note that a square root just means to take a number to the power of a half.

oh i get it

I did 1/2 (0.6021-1) instead of + because i did 40/100 and simplified it to 4/10 and separate that 2 logs as division which in logs is minus.

milkytea99 · Jul 2, 2015

Posted the solution. It's quite long but ah well lol

Log question (1 Viewer)

Green Yoda

Hi Φ

leehuan

Well-Known Member

Green Yoda

Hi Φ

milkytea99

Active Member

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