Question 1043188
(a)
{{{ log(6,(x+5))+ log(6,x)=2 }}}
The trick here is to express {{{ 2 }}} as a log base 6
{{{ 2 = log( 6, 36 ) }}}
Now you have:
{{{ log(6,(x+5))+ log(6,x)=log(6,36) }}}
{{{ log( 6, x*(x+5) ) = log( 6,36 ) }}}
{{{ x*( x+5 ) = 36 }}}
{{{ x^2 + 5x - 36 = 0 }}}
{{{ ( x + 9 )*( x - 4 ) = 0 }}} ( by looking at it )
{{{ x = 4 }}}
{{{ x = -9 }}}
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{{{ x = -9 }}} won't work as an answer because you get
{{{ log( 6, x ) = log( 6, -9 ) }}}
There is no way a real log to base 6 can give you a negative
number. Even negative logs give you positive answers.
So, {{{ x = 4 }}} is the answer
check:
{{{ log(6,(x+5))+ log(6,x)=2 }}}
{{{ log(6,(4+5))+ log(6,4)=2 }}}
{{{ log( 6,9 ) + log( 6,4 ) = 2 }}}
{{{ log( 6,36 ) = 2 }}}
{{{ 2 = 2 }}}
OK
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(b)
{{{ log(x) = log(2,8 ) }}}
You should know that {{{ log( 2,8 ) = 3 }}} so
{{{ log(x) = 3 }}}
The base is assumed to be {{{ 10 }}}, so
{{{ x = 10^3 }}}
{{{ x = 1000 }}}
OK