logarithms help?

Question

How do you solve: Use the properties of logarithims to simplify log 4 (4x^2) for x> 0?

Answer 1

Before you start the question, I'll state the log properties that may come into play:
1) log[base b](ac) = log[base b](a) + log[base b](c)

This basically states that the log of a product is the sums of the logs of its factors.

2) log[base b](a/c) = log[base b](a) - log[base b](c) for c non-zero.

This says the log of a quotient is the difference of the logs of the two numbers you are dividing.

3) log[base b](a^c) = c * log[base b](a)

This says that whenever you have a power inside of a log, you can take it outside of the log as a non-power.

Let's begin.

log[base 4](4x^2)

Apply log property #1, since 4x^2 = 4 * x^2

log[base 4](4x^2) = log[base 4](4) + log[base 4](x^2)

Now, remember that log[base b](b) is ALWAYS equal to 1 for b > 0. The reason why is because it translates to "b to the what power is b?" to which the answer is 1.

log[base 4](4x^2) = 1 + log[base 4](x^2)

Now, apply log property #3 to the existing log.

log[base 4](4x^2) = 1 + 2 log[base 4](x)

This should be in its reduced form.

Answer 2

hints first

log ( a x b ) = log a x log b
&
log ( a^b) = b x log (a)


answers

4 ( 4 x ^2 ) ; for x>0

simplifies to

16 x ^2

taking the log gives

log (16 x ^2) = log 16 + 2 log (x)

note that log 16 is a constant and your equation is a simple log x
why is this important?

because you can subtract log 16 from both sides, divide by 2, then take the inverse log of both sides to solve for x.

let's say for example

4 (4x^2) = y
16x^2 =y
log 16 + 2 log x = log y
2 log x = log y -log 16
log x = (log y - log 16) / 2

x = 10^ [ (log y - log 16)/2 ]

one other point, if x = 0, then 4 (4x^2) = 0 and taking the log of both sides gives log (0) on the right which makes your computer or calculator crash. why is that?

if x < 0, what happens?

Answer 3

hint: opposite powers by employing logarithms. So opposite a^n by taking the log of a^n with a base a; n=log[a](a^n) And opposite log base a by taking the skill of a. a^(log[a](n))=n.

Answer 4

Sry at first i didnot get what u typed!

I think its log [base 4] (4x^2)

Rules:

log [base a] b= log[base10] b/log[base10] a
Instead of 10 u can use any other number!

log(a*b) = log a + log b

log(a^b) = b*log a

log [base 4] (4x^2) = log (4x^2)/ log 4

= log {(2x)^2}/ log 4

= 2 log(2x) /2log 2

= log 2x/ log2

= log [base 2] 2x

_____ __________ _________ __________
My previous answer!!

I'm assuming base of every log in this problem to be 10

Properties of log to be used in this sum:
log(a*b) = log a + log b

log(a^b) = b*log a

log 4 (4x^2) for x> 0
= log 16*x^2
= log [(4*x)^2]
= 2*log [4*x]
= 2*(log 4 + log x)

Answer 5

log 4 (4x^2) = log4(4) + log4 (x^2) = 1 + 2log4 (x)