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Hello! I need some assistance with this homework question for precalculus, please?HW Q27

Hello I Need Some Assistance With This Homework Question For Precalculus PleaseHW Q27 class=

Sagot :

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.