evaluating logarithms...?

Using log laws we have that log (a)b = c => a^c = b

So, 5^x = 1/25 = 5^-2

=> x = -2

And, 27^x = 9

then, 3^3x = 9 = 3^2

i.e. x = 2/3

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Log5 1/25 =-2

5^-2 = 1/25

log is the inverse of raising a number to a power.

Log27 9 = 2/3

3^3 = 27, so 27^(1/3) = 3, 3^2 = 9, so [27^(1/3)]^2 = 9

That means 27^(2/3) = 9

1/25 = 5^-2 so log5 1/25 = -2

9 = 27 ^(2/3)

so log27 9 = 2/3

Log base 40 9 7 as an celebration is an same as 7 cases log base 40 9. Log base 25 to the ability of what's 5. The what's x. X is a variable, a range to sparkling up any equation. that is like this. Log 25 x= 25. So, what's x? X is two because 5 to the 0.33 power( 5 cases 5) is equivalent to25. Capiche? So extremely, that is 7 (log base 40 9)+ 27 (log base a million/9)/ (a million/32) (log base sixty 4)- (27/8) (log base 3/2)= the decision you are able to round it up or go away it as an same that is. it is the way you do it.

as by the logarithamic laws

log(a base b)= p => b=a^p

so by that

1/25=5^x = > 5^(-2) = 5^x => x= -2

similarly 27 = 9^x => 3^3=3^2x =>3=2x = > x = 1.5