Question 1031659
Use inverse functions
{{{ log( 5, 7x ) = 2 }}}
{{{ 5^2 = 7x }}}
{{{ 7x = 25 }}}
{{{ x = 25/7 }}}
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{{{ log( 3, n-3 ) = 4 }}}
{{{ 3^4 = n-3 }}}
{{{ n-3 = 81 }}}
{{{ n = 84 }}}
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{{{ log( x,49 ) = 2 }}}
{{{ x^2 = 49 }}}
{{{ x^2 = 7^2 }}}
{{{ x = sqrt( 7^2 ) }}}
{{{ x = 7 }}}
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{{{ log( 1000 ) = x }}}
( note that base 10 is assumed when
the base is not specified )
{{{ 10^x = 1000 }}}
{{{ 10^x = 10^3 }}}
{{{ x = 3 }}}
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{{{ log( 2, ((2^5)^x)^(-1) ) = 9 }}}
( hope I copied this right )
{{{ log( 2, 2^(-5x) )= 9 }}}
{{{ 2^9 = 2^( -5x ) }}}
{{{ 9 = -5x }}}
{{{ x = -9/5 }}}
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{{{ log( 4x - 1 ) = 2 }}}
{{{ 10^2 = 4x - 1 }}}
{{{ 4x = 10^2 + 1 }}}
{{{ 4x = 101 }}}
{{{ x = 25.25 }}}
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You may have trouble "reading" the
log functions
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In general, if you have:
log ( to some base ) ( that gives me this result ) = the log, which is an exponent
or, with symbols:
{{{ log( a, b ) = c }}}
The left side is TELLING you that the right side is a log ( exponent )
You know the base is {{{ a }}}, so the equation has to look like:
{{{ a^c = b }}}
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You can go in either direction, too
{{{ a^c = b }}}
{{{ log( a,b ) = c }}}
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Then you have to apply the laws of logs and
the inverse of the log -the exponent function
Hope this helps