Need help with logarithmic?

By revising some rules of logarithms , like such a rule:

log (a × b) = log a + log b

You can multiply the 2 by the 3 (=6) then by 3 again to get 18

so,

log 18 = log (2 × 3 × 3) = log 2 + log 3 + log 3 = a + b + b = a +2b

the answer, simply, is a +2b

Wish you the best :) and hope this helps

log 18 = log ( 2 x 3 ² ) = log 2 + log 3 ² = a + 2 b

assuming log base 10 although same for any base

given log 2 = a then by definition of logarithm 2 = 10^a

similarly if log3 = b => 3 = 10^b

18 = 2*3*3 = 10^a * 10^b * 10^b

18 = ( 10)^( a + 2b)

log(18)

log(2*3^2)

log(2) +log(3^2)

log(2) + 2log(3)

log(a) + 2log(b)

you could be tempted to do this

log(a) +log(b^2)

10^[log(a) + log(b^2)]

[10^log(a)][10^log(b^2)]

this is wrong, and you may not reduce it because you do not know what is it equal to. To convert to expoenents, you need to have a relation like an equals sign.

ab^2 is wrong

and

a +2b is so wrong its scary

There is a huge difference between and equality and an expression. all you can do with am expression is simplify, you can not solve.

by their nature, logs solve equalities by isolating exponents.

You are asked to express log(18) wrt a and b. we know what log 18 equal

log(18)=~1.255

ab^2=18

a+2b=8 these do not equal 1.255 so they are clearly wrong.

if you are allowed to introduce another variable, you could do something like

log(18)=log(x)

log(2*3^2)=log(x)

log(2)+log(3^2) = log(x)

log(a) + log(b^2) = log(x)

a+b^2 =x

what is very wrong is this

log(a) + 2log(b) =log(x)

a+2b =b

you can not drop the logs when they have a coeficient, that coeficieant is NOT a factor, its an exponent, and the expression looses all meaning if you treat like a factor.

log(ab)=loga+logb

log(18)

log(9*2)

log(3*3*2)

log3+log3+log2

b+b+a

2b+a