i need help with this log problem. 2logbase4(5x)=3

Try it first, then I'll help.

What's your 1st step?

well is it log4^3/2=5x?

Almost

But why's the log still there?

see here \[2\log_{4} (5x) \] is equal to 3

oh right. so it would be 4^3/2=5x and then i would have to figure what x is

Yep. So your answer is??

i'm not sure what to do with the fraction

now see according to log law \[k \log_{a} (n)=\log_{a}(n^k)\]

x^(a/b)=\[(\sqrt[b]{x})^{a}\]

how do i figure x with that?

I gave you a general formula. What is 4^(3/2) using that formula?

yes he is right ANNAZH don't be dependent on any1 fully now try it ur self any problem tell

do i cube or divide 5? im stuck

Leave 5 alone. Forget it for a second.

What is 4^(3/2)? From the formula I gave you above to solve any expression of the form x^(a/b), like this one.

would i changed radical 4 into 2 and then since it has the expo of 2, it is four?

So we have 4^(3/2)...take the square root of 4....and then raise it to the 3rd power. What do you get?

8?

Good. So 5x=8, and thus x=..?

8/5

ahh yes . thanks

\[log_b(a) = k \iff b^k = a\] Also \[a\cdot log (b) = log(b^a)\] Therefore \[2log_4(5x)=3\]\[\implies log_4([5x]^2) = 3\]\[\implies log_4(25x^2) = 3\]\[\implies 25x^2 = 4^3\]\[\implies x = \sqrt{4^3 \over 25}\]\[\implies x = {2^3 \over 5} = {8 \over 5}\]

Nicely done both of you.

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