Simplifying logarithmic expressions

[fatima_a]

simplify the following expression

log2 4^3x-1 (2 is a subscript of log and 3x - 1 is a superscript of 4)

i can rearrange this to 4log2 3x - 1,
i am thinking after this it becomes 4 log3 + 4 logx, but i don't know what to do with the -1

please show all the steps and explain.

 

[LCKurtz]

You need to use parentheses to keep your variables straight. And use the Go Advanced button so you have access to the X² and X₂ superscript and subscript buttons. Using these buttons, you expression looks like this: log₂ 4^(3x-1)

Write 4 as 2² and use the property of logs that says:

log_a(a^b) = b

and don't lose track of your parentheses.

 

[HallsofIvy]

simplify the following expression

log2 4^3x-1 (2 is a subscript of log and 3x - 1 is a superscript of 4)

i can rearrange this to 4log2 3x - 1,

You appear to be thinking that "$log(b^x)= b log(x)$". That is wrong. The correct formula is [itex]log(b^x)= x log(b)$.<br /> <br /> <blockquote data-attributes="" data-quote="" data-source="" class="bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch"> <div class="bbCodeBlock-content"> <div class="bbCodeBlock-expandContent js-expandContent "> i am thinking after this it becomes 4 log3 + 4 logx, but i don't know what to do with the -1 </div> </div> </blockquote> No, "log (a+ b)" is NOT equal to "log(a)+ log(b)" but that is no longer relevant.<br /> <br /> <blockquote data-attributes="" data-quote="" data-source="" class="bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch"> <div class="bbCodeBlock-content"> <div class="bbCodeBlock-expandContent js-expandContent "> please show all the steps and explain. </div> </div> </blockquote> Go back and review the "laws of logarithms".$[/itex]