logarithms: solve 2log_3(4) + log_3(X) = 5log_3(2)

[hlpmeplease]

2log[sub:2aycet1j]3[/sub:2aycet1j]4 + log[sub:2aycet1j]3[/sub:2aycet1j]X = 5log[sub:2aycet1j]3[/sub:2aycet1j]2

Solve for x.

I do know that: Log a + Log b = log ab, but for some reason I keep freezing up on this problem. Im not quite sure how to start out.

 

[hlpmeplease]

Re: logarithms

Would this start be correct?
Log[sub:2j8vwiya]3[/sub:2j8vwiya]4[sup:2j8vwiya]2[/sup:2j8vwiya]X =Log[sub:2j8vwiya]3[/sub:2j8vwiya]2[sup:2j8vwiya]5[/sup:2j8vwiya]

 

[Deleted member 4993]

Re: logarithms

Would this start be correct?
Log[sub:1e56ae0q]3[/sub:1e56ae0q]4[sup:1e56ae0q]2[/sup:1e56ae0q]X =Log[sub:1e56ae0q]3[/sub:1e56ae0q]2[sup:1e56ae0q]5[/sup:1e56ae0q] <<< Correct

 

[stapel]

Would this start be correct?
Log[sub:egd0wynl]3[/sub:egd0wynl]4[sup:egd0wynl]2[/sup:egd0wynl]X =Log[sub:egd0wynl]3[/sub:egd0wynl]2[sup:egd0wynl]5[/sup:egd0wynl]

Exactly! Then, since you've got "log(something) = log(something else)", equate the (something) and the (something else), and solve the resulting quadratic equation. :wink:

 

[Deleted member 4993]

Stapels advice in general is correct. However, in this case the resulting equation is linear.

(something) = (something else)

4[sup:1u5zybri]2[/sup:1u5zybri]X = 2[sup:1u5zybri]5[/sup:1u5zybri]

2[sup:1u5zybri]4[/sup:1u5zybri]X = 2[sup:1u5zybri]5[/sup:1u5zybri]

X = 2