logarithmic functions (problem inside)

- katherinesmith

logarithmic functions (problem inside)

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- katherinesmith

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

- anonymous

hug me?

- phi

is the -5 part of the log ?

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## More answers

- katherinesmith

there are no parenthesis

- phi

without parens, the -5 is outside the log

- katherinesmith

really. weird

- katherinesmith

i have no idea how to solve that

- phi

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

- katherinesmith

x = 1/2 ?

- phi

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

- katherinesmith

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

- phi

divide both sides by 3
take the log of both sides
use the property that
\[\log(a^b) = b \log(a) \]

- katherinesmith

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

- katherinesmith

x = 1.8413

- phi

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

- katherinesmith

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

- phi

I think you did 6/3 wrong

- katherinesmith

6/3 = 2

- phi

now keep going

- katherinesmith

OH WAIT

- katherinesmith

i figured it out

- katherinesmith

x = 2 ?

- phi

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

- phi

somehow you got ln(3)/ln(5) ...
\[ x= \frac{3}{2}+ \frac{\ln2}{2 \ln5} \]

- katherinesmith

is that the final answer?

- phi

you could change it to a decimal with a calculator

- katherinesmith

this hurts my brain

- katherinesmith

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

- phi

always combine like terms... the 10 and the 22
add -10 to both sides
now it is the same problem as the previous one

- katherinesmith

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

- phi

so far you have
2^(x-4) = 12
to undo the exponent , use a log (or ln) on both sides

- katherinesmith

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

- katherinesmith

am i on the right track

- phi

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

- katherinesmith

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

- phi

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

- katherinesmith

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

- phi

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

- katherinesmith

i have one left. will open a new question

- phi

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....

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