Question 405740

{{{log (8x + 15)=1}}} ....base is {{{10}}}, so

 {{{(8x + 15)=10^1}}}

 {{{8x+ 15=10}}}

{{{8x=10-15}}}

{{{x=-5/8 = -0.625}}}



{{{8x + 15=8*(-0.625) +15= 5+15=10}}}......so, {{{log (10)=1}}}