If x is a positive real number different from 1 such that logax, logbx, logcx are in A.P., then

Question

If x is a positive real number different from 1 such that logax, logbx, logcx are in A.P., then (A) b=(a+c/2) (B) b=√ac (C) c2=(ac)loga b (D) None of (A), (B), (C) are correct

Tardigrade Admin · Accepted Answer

2logbx=logax+logcx=(1/logxa)+(1/logxc) ⇒ (2/logxb)=(logxac/logxa logxc) ⇒ 2 logxc=(logxb/logxa)(logx ac)=logab.logxac ⇒ c2=(ac)loga b

Q. If $x$ is a positive real number different from $1$ such that $l o g_{a} x, l o g_{b} x, l o g_{c} x$ are in A.P., then

$b = \frac{a + c}{2}$

$b = a c$

$c^{2} = (a c)^{l o g_{a} b}$

None of (A), (B), (C) are correct

Q. If x is a positive real number different from 1 such that loga​x,logb​x,logc​x are in A.P., then

b=2a+c​

b=ac​

c2=(ac)loga​b

None of (A), (B), (C) are correct

2logb​x=loga​x+logc​x=logx​a1​+logx​c1​⇒logx​b2​=logx​alogx​clogx​ac​ ⇒2logx​c=logx​alogx​b​(logx​ac)=loga​b.logx​ac ⇒c2=(ac)loga​b

Q. If $x$ is a positive real number different from $1$ such that $l o g_{a} x, l o g_{b} x, l o g_{c} x$ are in A.P., then

$b = \frac{a + c}{2}$

$b = a c$

$c^{2} = (a c)^{l o g_{a} b}$