katherinesmith
  • katherinesmith
log5x = 3
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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katherinesmith
  • katherinesmith
\[\log _{5}x = 3\]
Luigi0210
  • Luigi0210
You could start by converting it to an exponent equation
katherinesmith
  • katherinesmith
notice the 5 is a subscript

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Luigi0210
  • Luigi0210
Yes, I see that. You can still convert it.
katherinesmith
  • katherinesmith
okay, how.
phi
  • phi
make each side the exponent of 5
phi
  • phi
making the log_5 the exponent of the base (5 here) "undoes" the log \[ b^{\log_b(a)} = a \]
katherinesmith
  • katherinesmith
can you plug the numbers into the start of the equation?
phi
  • phi
you start with \[ \log _{5}x = 3 \] make each side the exponent of the base = 5 \[ 5^{\log _{5}x} = 5^3\]
katherinesmith
  • katherinesmith
okay...
phi
  • phi
the ugly expression on the left side is just x (which is why we did it)
katherinesmith
  • katherinesmith
x = 25?
phi
  • phi
x= 5^3 = 5*5*5= 125
katherinesmith
  • katherinesmith
that 3 looks like a 2 because of how small the font is
katherinesmith
  • katherinesmith
logarithms make me want to die slowly
phi
  • phi
logs are confusing. but logs and exponents of a base go together.
katherinesmith
  • katherinesmith
\[\log _{2} (x - 3) = 5\]
phi
  • phi
same idea, different base
katherinesmith
  • katherinesmith
same formula?
phi
  • phi
\[ b^{\log_b(a)} = a \]
katherinesmith
  • katherinesmith
\[2^{\log2(x - 3)} = 2^{5}\]
katherinesmith
  • katherinesmith
is that right so far
phi
  • phi
as long as the big 2 is really a little 2 in the lower right.
katherinesmith
  • katherinesmith
the big two in the front or the end?
phi
  • phi
the log_2 should give you a 2 in the lower right
phi
  • phi
\[ 2^{\log_2(x - 3)} = 2^{5} \]
katherinesmith
  • katherinesmith
okay so then you have x(x-3) = 2^5 correct?
phi
  • phi
just (x-3) = 2^5
katherinesmith
  • katherinesmith
x = 35
phi
  • phi
this may be confusing, but it should be easy to remember \[ \cancel{2}^{\cancel{\log_2} x} = x \]
katherinesmith
  • katherinesmith
\[3\log _{6} (x + 1) = 9\]
katherinesmith
  • katherinesmith
is my next one
phi
  • phi
divide both sides by 3 then you have the same problem, with a different base
katherinesmith
  • katherinesmith
\[\log _{6}(x + 1) = 3\]
katherinesmith
  • katherinesmith
i got x = 215
phi
  • phi
yes
katherinesmith
  • katherinesmith
woohooo! i will open a new question

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