General Algerbra - ln and log

Rewrite ln(e^2x+1)

ln and e cancel out

2x + 1

Rewrite 7ln(e^(2x+1))

ln and e cancel out

7(2x+1)
which becomes
14x + 7

Rewrite ln(2xcos(x))

RULE : ln(ab) = ln(a) + ln(b)
so
ln(2x) + ln(cos(x))
and again
ln(2) + ln(x) + ln(cos(x))

cos(x) is a number NOT cos multiplied by x so we leave it there

Rewrite ln(2x/7)

RULE : ln(a/b) = ln(a) - ln(b)

ln(2x) - ln(7)

Rewrite ln(x⁴)

RULE : ln(a^r)=rln(a)

ln(x⁴) = 4ln(x)

What is e⁰

1

anything to the power of 0 is 1

What is ln(1)

0

What is ln(e)

1

Rewrite ln(a+b)

Trick question, you cannot!

There is no rule for splitting addition or subtraction inside ln

Rewrite ln(a-b)

Trick question, you cannot!

There is no rule for splitting addition or subtraction inside ln

Rewrite e^{ln(2x + x)}

ln and e cancel out

3x

Did you forget to combine the x’s? You need to remember to combine like terms!

Rewrite log₃(2x) in terms of ln

RULE : log_b(z) = ln(z) / ln(b)

so
ln(2x) / ln(3)

Rewrite log₇(xy)

Rule: log_c(ab) = : log_c(a) + log_c(b)

log₇(x) + log₇(y)

Rewrite log(3x)

Rule: log_c(ab) = : log_c(a) + log_c(b)

log(3) + log(x)

Rewrite log(3/x³)

Rule: log_c(a/b) = : log_c(a) - log_c(b)

log(3) - log(x³)

Rewrite log₁₂(10/6x)

Rule: log_c(a/b) = : log_c(a) - log_c(b)

log₁₂(10) - log₁₂(6x)

Rewrite log(x³)

Rule: log_b(x^r) = rlog_b(x)

3log(x)

Just like ln and e, logs and powers cancel each other out.

Tell me two examples

This only applies when the base of the power and the base of the log are the same

2^log₂(13x) can be rewritten as : 13x

AND

log₁₅(15^2x+2) can be rewritten as : 2x + 2

Does the log and the power cancel out in

log₅(x⁵)

NO

The base of the log is 5 and the base of the power is x.

They do not equal so they do not cancel!!

Rewrite log₁₂(1)

0

Whenever there is only 1 in the brackets it is 0

Rewrite log_x(1)

0

Whenever there is only 1 in the brackets it is 0

Rewrite log₁₂(12)

1

when ever the number in the brackets equals the base of the log it is 1

Rewrite log₆(1+5)

1

when ever the number in the brackets equals the base of the log it is 1

Rewrite log₃(√23)

RULE: log_b(ⁿ√x) = 1/n log_b(x)

so, since the question is the square root ( √23 is the same as writing ²√23)
(1/2)log₃(23)

Rewrite log₅(⁶√2)

RULE: log_b(ⁿ√x) = 1/n log_b(x)

so
(1/6)log₅(2)