Question 1149864
(1) {{{ log( 5, (x+2) ) = 1 - log( 5, ( x+6 ) ) }}}
(1) {{{ log( 5, (x+2) ) = log( 5,5 ) - log( 5, ( x+6 ) ) }}}
(1) {{{ log( 5, (x+2) ) = log( 5, ( 5/( x+6 )) ) }}}
(1) {{{ x + 2 = 5/( x+6 ) }}}
(1) {{{ ( x+2 )*( x+6 ) = 5 }}}
(1) {{{ x^2 + 8x + 12 = 5 }}}
(1) {{{ x^2 + 8x + 7 = 0 }}}
(1) {{{ ( x + 7 )*( x + 1 ) = 0 }}}
{{{ x = -7 }}}
{{{ x = -1 }}}
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check answer:
(1) {{{ log( 5, (x+2) ) = 1 - log( 5, ( x+6 ) ) }}}
(1) {{{ log( 5, (-1+2) ) = 1 - log( 5, ( -1+6 ) ) }}}
(1) {{{ log( 5, 1 ) = 1 - log( 5, 5 ) }}}
(1) {{{ 0 = 1 - 1 }}}
OK
Note that {{{ x = -7 }}} does not work because
it gives you:
(1) {{{ log( 5, (-7+2) ) = 1 - log( 5, ( -7+6 ) ) }}}
There is no log to a positive base that gives a
negative result
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(2) {{{ log( 2, (x+7) ) = 2 + log( 2, (x+4) ) }}}
(2) {{{ log( 2, (x+7) ) = log( 2, 4 ) + log( 2, (x+4) ) }}}
(2) {{{ log( 2, (x+7) ) = log( 2, 4*( x+4 ) ) }}}
(2) {{{ x + 7 = 4*( x+4 ) }}}
(2) {{{ x + 7 = 4x + 16 }}}
(2) {{{ 3x = -9 }}}
(2) {{{ x = -3 }}}
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check answer:
(2) {{{ log( 2, (x+7) ) = 2 + log( 2, (x+4) ) }}}
(2) {{{ log( 2, (-3+7) ) = 2 + log( 2, (-3+4) ) }}}
(2) {{{ log( 2, 4 ) = 2 + log( 2, 1 ) }}}
(2) {{{ 2 = 2 + 0 }}}
OK