2logx to the base 3 - 3log4 to the base 2 = log1 to the base b

Question

solomonzelman · Answer

What is the question?

solomonzelman · Answer

Oh the b?

anonymous · Answer

Solve for x

campbell_st · Answer

is this the question?

$$2\log_{3}(x) - 3\log_{2}(4) = \log_{b}(1)$$

anonymous · Answer

Yes... That's the question thank you

campbell_st · Answer

ok... so 

$$\log_{2}(4) = \log_{2}(2^2) = 2$$

so you have 

$$\log_{3}(x^2) - 6 = \log_{b}(1)$$

so I'd suggest using change of base...

anonymous · Answer

How would I do that? I also get to that step but can't get further

campbell_st · Answer

ok... so this is how I would do the question, after re-reading it

the log of 1 to any base is always zero... its just a fact.   

and you saw how I managed the base 2 log of 4

so you have 

$$2\log_{3}(x) - 6 = 0$$

which becomes 

$$2 \log_{3}(x) = 6$$

divide both sides of the equation by 2

$$\log_{3}(x) = 3$$

now raise each side of the equation to the power of 3 

$$3^{\log_{3}(x)} = 3^3$$

which is 

x = 27

anonymous · Answer

Thank you so much