Question 211487
problem is:
{{{log(x,root(3,9)) = 1/6}}}
by the definition of logarithms, this equation is true if and only if:
{{{x^(1/6) = root(3,9)}}}
that equation is the same as:
{{{x^(1/6) = 9^(1/3)}}}
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take the 6th power of both sides of this equation to get:
{{{(x^(1/6))^6 = (9^(1/3))^6}}}
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by the laws of exponents, {{{(x^a)^b = x^(a*b)}}} so the equation of {{{(x^(1/6))^6 = (9^(1/3))^6}}} is equivalent to:
{{{x = 9^2}}}
this makes x = 81
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let's see if this holds water.
the original equation is:
{{{log(x,root(3,9)) = 1/6}}}
we replace x with 81 to get:
{{{log(81,root(3,9)) = 1/6}}}
this is true if and only if {{{81^(1/6) = root(3,9)}}}
using the calculator, {{{81^(1/6)}}} = 2.080083823...
using the calculator {{{root(3,9)}}} = cube root of 9 = 2.080083823...
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looks like they're equivalent.
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answer is x = 81
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you can also solve this using your calculator by converting from the log to the base of 81, to the log to the base 10 which your calculator can handle.
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{{{log(81,root(3,9)) = 1/6}}}
using the logarithm base conversion formula,
{{{log(81,root(3,9)) = (log(10,root(3,9)))/(log(10,81))}}}
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using the calculator,
(log(10,root(3,9)))/(log(10,81))}}} becomes:
.318080836/1.908485019 = .16666666...
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This is equivalent to 1/6 confirming that the answer is correct.
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answer is:
x = 81
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