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

Question

Sophie Adams · Accepted Answer

# x=1/6# #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.

Serenity Johnson · Answer

#"see 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#

Elijah Allen · Answer

We will start the proof

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#

Christian Antunez · Answer

undefined To solve this problem, we'll use the properties of logarithms.

Given:
1. $ \log_{b}a = \frac{1}{x} $
2. $ \log_{a}\sqrt{b} = 3x^2 $

We'll start by expressing $ \log_{a}\sqrt{b} $ as $ \frac{1}{2}\log_{a}b $, since $ \sqrt{b} = b^{\frac{1}{2}} $.

So, $ \frac{1}{2}\log_{a}b = 3x^2 $

Multiplying both sides by 2, we get:
$ \log_{a}b = 6x^2 $

Now, we can use the given $ \log_{b}a = \frac{1}{x} $ to rewrite $ \log_{a}b $ as $ \frac{1}{x} $:

$ \frac{1}{x} = 6x^2 $

Solving for $ x $:

$ x = \frac{1}{\sqrt{6}} $

Thus, $ x = \frac{1}{6} $.