Question

How to solve this problem concerning logarithms?

a, b, c, d are positive real numbers, so that `log_a(b)=3/2` and `log_c(d)=5/4`.
If a-c=9, what is the solution of b-d?

Answer 1

Given:

`log_a(b)=3/2" [1]"`

`log_c(d)=5/4" [2]"`

Raise both sides of both equation to their respective bases:

`b = a^(3/2)" [1.1]"`

`d= c^(5/4)" [2.1]"`

Substitute `a = c+9` into equation [1.1]:

`b = (c+9)^(3/2)" [1.2]"`

`d= c^(5/4)" [2.1]"`

Subtract equation [2.1] from equation [1.2]:

`b-d = (c+9)^(3/2)- c^(5/4)`

Answer 2

(see Douglas K's answer for the question as stated, however...)
If `a,b,c,d` are positive Integers then `b-d=93`

Explanation:

`log_a b=3/2 rarr a^(3/2) = b (in ZZ^+)`
and if `b in ZZ^+ rArr a` is the square of an integer i.e. `a=p^2`, for some `p in ZZ^+`

Similarly, `log_c d=5/4rarr c^(5/4)=d (n ZZ^+)`
and if `d inn ZZ^+ rArr c` is the square of the square of an integer i.e. `c=q^4=(q^2)^2` for some `q in ZZ+`

We are given that `a-c=9`
`rArr p^2-(q^2)^2=9`

`rarr (p-q^2)(p+q^2)=9`
`color(white)("XXX")`Note: `p,q in ZZ^+ rarr (p+q^2) in ZZ+ rarr (p-q^2) in ZZ+`
`color(white)("XXXXXX")` and `(p+q^2) > (p-q^2)`

If `(p+q^2)` and `(p-q^2)` are integer factors of `9`
then the only possibility is
`color(white)("and")p+q^2=9`
and `p-q^2=1`
#{: ("Adding:",2p=10,color(white)("xxx")"Subtracting:",2q^2=8), (,p=5,,q=2) :}#

#{: ("Since:", a=p^2,color(white)("xxx")"and since:", c=q^4), (,a=25,,c=16) :}#

and from there:
#{: ("Since:", b=a^(3/2),color(white)("xxx")"and since:", d=c^(5/4)), (,b=125,,d=32) :}#

and `b-d=93`