If #log_ba=1/x and log_a √b =3x^2#, show that x =1/6?

3 Answers
Jim G.
Jun 3, 2018

#"see explanation"#

Explanation:

#"using the "color(blue)"law of logarithms"#

#•color(white)(x)log_b x=nhArrx=b^n#

#log_b a=1/xrArra=b^(1/x)#

#log_a b^(1/2)=3x^2rArrb^(1/2)=a^(3x^2)#

#"substitute "a=b^(1/x)" into "a^(3x^2)#

#b^(1/2)=(b^(1/x))^(3x^2)=b^(3x)#

#"we have "b^(1/2)=b^(3x)#

#3x=1/2#

#"divide both sides by 3"#

#x=(1/2)/3=1/6#

clarrisse · Stefan V.
Jun 3, 2018

# x=1/6#

Explanation:

#log_b a=1/x#

So

#xlog_ba=1#

#x=1/log_b a#

#x=log_ab#

Now

#1/2log_a b=3x^2" "# (substituting from #log_a b=x#)

So

# x/2=3x^2#

# x=6x^2#

# 6x=1#

# x=1/6# with #x !=0#

Hence proved.

Sonnhard
Jun 3, 2018

We will start the proof

Explanation:

We have
#ln(a)/ln(b)=1/x#
and

#1/2*ln(b)/ln(a)=3x^2#
so

#ln(a)/ln(b)=1/(6x^2)#

using the first equation we get
#1/x=1/(6x^2)#

multiplying by #x^2# we get

#x=1/6# for #xne 0#

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