Question

logarithmic functions (problem inside)

Answer 1

\[\log _{5}2x - 5 = -4\]

Answer 2

hug me?

Answer 3

is the -5 part of the log ?

Answer 4

there are no parenthesis

Answer 5

without parens, the -5 is outside the log

Answer 6

really. weird

Answer 7

i have no idea how to solve that

Answer 8

add 5 to both sides
then it is the same problem with a different base

Answer 9

x = 1/2 ?

Answer 10

no wait. you should get log_5(2x) = 1
make each side the exponent of base 5:

5^log_5(2x) = 5^1
2x = 5
x= 5/2

Answer 11

what about ones like this
\[3(5)^{2x - 3} = 6\]

Answer 12

divide both sides by 3
take the log of both sides

use the property that
\[\log(a^b) = b \log(a) \]

Answer 13

\[x = \frac{ 3 }{ 2 } + \frac{ \ln (3) }{ 2\ln (5) }\]

Answer 14

x = 1.8413

Answer 15

go in smaller steps... you are close, but you went astray

Answer 16

i don't know what to do differently! i thought that was right

Answer 17

I think you did 6/3 wrong

Answer 18

6/3 = 2

Answer 19

now keep going

Answer 20

OH WAIT

Answer 21

i figured it out

Answer 22

x = 2 ?

Answer 23

\[ 3(5)^{2x - 3} = 6 \\(5)^{2x - 3} = 2 \\ \ln\left(5^{2x - 3}\right)= \ln2 \\ (2x-3) \ln(5)= \ln(2) \\2x-3 = \frac{\ln(2)}{\ln(5)}\]

Answer 24

somehow you got ln(3)/ln(5) ...

\[ x= \frac{3}{2}+ \frac{\ln2}{2 \ln5} \]

Answer 25

is that the final answer?

Answer 26

you could change it to a decimal with a calculator

Answer 27

this hurts my brain

Answer 28

\[2^{x - 4} + 10 = 22\]

Answer 29

always combine like terms... the 10 and the 22
add -10 to both sides

now it is the same problem as the previous one

Answer 30

and i have no idea how i solved the previous one. ugh

Answer 31

so far you have
2^(x-4) = 12

to undo the exponent , use a log (or ln) on both sides

Answer 32

\[x - 4 = \frac{ \ln 12 }{ \ln 2 }\]

Answer 33

am i on the right track

Answer 34

yes. though you did a few steps all at once... if you really are unsure, go slower.

Answer 35

i wrote them out on paper and just showed you the last step i reached because now i'm not sure what to do

Answer 36

you are solving for x. you want x by itself on the left side

Answer 37

\[x = 4 + \frac{ \ln 12 }{ 2\ln 2 }\]

Answer 38

yes. if you use a calculator to find x, and use that value in the original equation , it should check out.

Answer 39

i have one left. will open a new question

Answer 40

one problem... you had
\[ x - 4 = \frac{ \ln 12 }{ \ln 2 } \]
you should get
\[ x = 4+ \frac{ \ln 12 }{ \ln 2 } \]

somehow you got 2 ln(2) in the bottom....