Hello! I need some assistance with this homework question for precalculus, please?HW Q27
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The expression is given to be:
[tex]\log_55^{86}[/tex]Recall the rule of logarithms:
[tex]\log a^b=b\log a[/tex]Thus the expression becomes:
[tex]\log_55^{86}=86\log_55[/tex]Recall the rule:
[tex]\log_aa=1[/tex]Hence, we have the expression to be:
[tex]86\log_55=86[/tex]The answer is 86.