Maths- exponent question

Question

If 3^a = 21^c and 7^d=21^c

show that c= ad/a+d

Answer 1

there may be an error in the problem,

a problem with the problem

a miscopy or misprint somewhere

an integer or two off a little

maybe 9 instead of 7

27 instead of 21

useful log property is

logab = loga + logb

such as

log21 = log3 + log7

if log base =3

then log3 = 1

log21 = 1 +log7 = 1 +ln7/ln3 with log base =3

as is,

3^a=7^d

a=dln7/ln3

=1.771243749

=about 1.77d

a=3c

c=a/3= .590414583

c=about .59d

ad/(a+d)=1.77d^2/(2.77d)

=1.77d/2.77=.64d

.59d doesn't = .64d,

close, but one counter example disproves the "identity"

that c = ad/(a+d)

21^c= (3x7)^c= (3^c)(7^c)= (7^d)

3^a=(7^d)= 21^c=(3^c)(7^c)

c= log ₂₁(3^a) =log₂₁ (7^d)= log₂₁[(3^c)(7^c)]

c=ad/(a+d)

a=log₃21^c=clog₃21=3c

d=log₇21^c=clog₇21=c(log₇7+log₇3)=c+clog₇3= c+cln3/ln7

c= ad/(a+d)= (3cd)/(3c+d)=[(3c^2+3c^2(ln3/ln7)]/(3c+cln3/ln7)

=[3c+3cln3/ln7)/(3+ln3/ln7)

=c(3+3ln3/ln7)(3+ln3/ln7)

right side is close to c(1) but not quite, it needs another 3 in the denominator's 2nd term

try

3^a = 9^d = 28^c

with a=4, d=2, c= 4/3

then

c = ad/(a+d) = 8/6 = 43/

3^4=9^2=28^(4/3) = 81

a sloppy 9 can look like 7

a sloppy 7 can look like 1

27 confused with 21

27^c gets written as 21^c

9^d as 9^7