How to solve log(base 2)(24) - log(base 2)(3) = log(base 5)(x)?

Question

Okay, here's the equation:

log(base 2)(24) - log(base 2)(3) = log(base 5)(x).

We're solving for x here. I get that you use the quotient rule.

Then you get:
log(base 2)(24/3) = log(base 5)(x)
log(base 2)(8) = log(base 5)(x)

But what do you do after that point and WHY do you do it (i.e. what rules say to do it that way)? THANK YOU SO MUCH!!! I'll pick a best answer TODAY!

Answer 1

the best way would be to use the change of base rule. The change of base rule basically works by changing the two different numbers to the same bases logarithm by division, like this:
log(base2) 3 = log(base10)3 / log(base10) 2

the neat thing is the new base can be whatever number you want (which is useful when working with natural logs of base e).

In this problem, you can also simplify log(base2) 8 = 3, which may make things a little easier.

So you can work this problem like this (I'm going to assume log8 is the same as log(base10)8):

log(base2) 8 = log(base5) x
3 = logx / log5
logx = 3 log5
logx = log(5^3)
x = 5^3
x = 125

Answer 2

u need to get rid of the logs by raising the log base 2 to the base 2 and the log base 5 to the base 5.
i.e. it becomes 2^8=5^x
therefore 256=5^x
there x=log(base5)256

it isnt really using rules as such its more knowing that e^2 and log(base2) are opposites of each other and will cancel out.

Answer 3

log[3] (log[x] (log[4] 16)) = -1    → (log[4] 16) = 2, therefore log[3] (log[x] 2) = -1 In the exponential form, this would be written log [x] 2 = 3ˉ¹, or log [x]2 = ⅓ and again, putting this into exponential form x^⅓ = 2 Cube both sides x = 2³ x = 8

Answer 4

LHS
log means log to base 2 at this stage:-
log 24 - log 3 = log (24/3) = log 8 = 3

RHS
log now taken to mean log to base 5
So have log x

But LHS = RHS
3 = log x (where log is to base 5)
x = 5³
x = 125

Answer 5

log(base 2)(8)=log(base 5)(x)
2^[log(base 5)(x)]=8
log(base 5)(x) ln(2) = ln(8)
log(base 5)(x) = ln(8) / ln(2)
x= 5^[ln(8) / ln(2)]
ln is log to the base e.
Alternatively, you can use log(base 10) instead of log (base e).