log4 64 +log4 4

Question

log4 64 +log4 4

anonymous · Answer

$$\log_{4}64+\log_{4}4  $$

jhannybean · Answer

$$\log(a) + \log(b) = \log(ab)$$

jhannybean · Answer

So $$\log_4 (64) + \log_4 (4) = p$$$$\log_4 (4\cdot 64) = p$$$$\log_a (b) = x \implies a^x = b$$$$4^p = (4 \cdot 64)$$

jhannybean · Answer

Now what is 4 x 64?

anonymous · Answer

256

jhannybean · Answer

What power do you need to raise 4 to to get 256?

anonymous · Answer

16

jhannybean · Answer

$$\large 4^{16} \ne 256$$

jhannybean · Answer

Think much much much smaller.

anonymous · Answer

sorry    4

jhannybean · Answer

Good. $$4^4 = 256$$Now we replace 256 in our equation. $$4^p = 4^4$$ notice now they have the same base?

anonymous · Answer

ya

jhannybean · Answer

$$\color{red}4^p = \color{red}4^4$$$\color{red}4$ is the base.

jhannybean · Answer

Ok, we are trying to evaluate their powers, or rather, solve for p, which is a variable constant.

In order to do that, we need to take $\log_4$ of the function on both sides.

jhannybean · Answer

So $$\log_4 4^p = \log_4 4^4$$$$\log_4(4) = 1$$ So what would be our answer?

anonymous · Answer

does 4 cancel?

jhannybean · Answer

Yes, when you take the log base of a number that is the same for the function you are evaluating, the log and function cancel out.

anonymous · Answer

so its 1?

jhannybean · Answer

Sorry, perhaps I did not clarify enough.

jhannybean · Answer

when you take the log function of both sides, the exponent of the function comes out in front. $$\log_4 (4^p) = \log_4 (4^4)$$$$(p)\log_4(4) =(4)\log_4(4)$$

jhannybean · Answer

So what does $\log_4(4)$ become?

anonymous · Answer

sorry im lost I havnt really learned trig but I have a state exam tom

jhannybean · Answer

What portion are you confused about? Maybe I can clarify it for you.

anonymous · Answer

after 4^4=256

jhannybean · Answer

So we replaced the 256 with $4^4$ into our equation. did you understand that?

anonymous · Answer

ya

jhannybean · Answer

ok, we rewrote it as: $4^p = 4^4$ Now we want to evaluate (solve) for whats in the exponents, but we need to get rid of the base $4$

mathstudent55 · Answer

Just because a rule exists, it does not mean it must be used.
here is a rule you can use in this problem: $\log_b b^n = n$

$\log_{4}64+\log_{4}4$

$= \log_4 4^3 + \log_4 4^1$

$ = 3 + 1$

$= 4$

jhannybean · Answer

Oh, thanks for that @mathstudent55 :)