katherinesmith
  • katherinesmith
logarithmic functions (problem inside)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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katherinesmith
  • katherinesmith
\[\log _{5}2x - 5 = -4\]
anonymous
  • anonymous
hug me?
phi
  • phi
is the -5 part of the log ?

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More answers

katherinesmith
  • katherinesmith
there are no parenthesis
phi
  • phi
without parens, the -5 is outside the log
katherinesmith
  • katherinesmith
really. weird
katherinesmith
  • katherinesmith
i have no idea how to solve that
phi
  • phi
add 5 to both sides then it is the same problem with a different base
katherinesmith
  • katherinesmith
x = 1/2 ?
phi
  • 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
  • katherinesmith
what about ones like this \[3(5)^{2x - 3} = 6\]
phi
  • phi
divide both sides by 3 take the log of both sides use the property that \[\log(a^b) = b \log(a) \]
katherinesmith
  • katherinesmith
\[x = \frac{ 3 }{ 2 } + \frac{ \ln (3) }{ 2\ln (5) }\]
katherinesmith
  • katherinesmith
x = 1.8413
phi
  • phi
go in smaller steps... you are close, but you went astray
katherinesmith
  • katherinesmith
i don't know what to do differently! i thought that was right
phi
  • phi
I think you did 6/3 wrong
katherinesmith
  • katherinesmith
6/3 = 2
phi
  • phi
now keep going
katherinesmith
  • katherinesmith
OH WAIT
katherinesmith
  • katherinesmith
i figured it out
katherinesmith
  • katherinesmith
x = 2 ?
phi
  • 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
  • phi
somehow you got ln(3)/ln(5) ... \[ x= \frac{3}{2}+ \frac{\ln2}{2 \ln5} \]
katherinesmith
  • katherinesmith
is that the final answer?
phi
  • phi
you could change it to a decimal with a calculator
katherinesmith
  • katherinesmith
this hurts my brain
katherinesmith
  • katherinesmith
\[2^{x - 4} + 10 = 22\]
phi
  • 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
  • katherinesmith
and i have no idea how i solved the previous one. ugh
phi
  • phi
so far you have 2^(x-4) = 12 to undo the exponent , use a log (or ln) on both sides
katherinesmith
  • katherinesmith
\[x - 4 = \frac{ \ln 12 }{ \ln 2 }\]
katherinesmith
  • katherinesmith
am i on the right track
phi
  • phi
yes. though you did a few steps all at once... if you really are unsure, go slower.
katherinesmith
  • 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
  • phi
you are solving for x. you want x by itself on the left side
katherinesmith
  • katherinesmith
\[x = 4 + \frac{ \ln 12 }{ 2\ln 2 }\]
phi
  • phi
yes. if you use a calculator to find x, and use that value in the original equation , it should check out.
katherinesmith
  • katherinesmith
i have one left. will open a new question
phi
  • 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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