Calian wants to construct a viable argument to justify that logbb=1

Question

log
𝑏
⁡
𝑏
=
1
for b>0
𝑏
>
0
and b≠1
𝑏
≠
1
. Fill in the blanks to help him construct his argument.(4 points)
Any positive number raised to the power is equal to itself. Written as an equation, this means that $$b=b​​​​​​​ for any positive base. This equation is equivalent to the logarithmic equation $$log​b=. Therefore, it must be true that logbb=1
log
𝑏
⁡
𝑏
=
1
for any positive base that is not equal to 1.

Answer 1

To construct a viable argument to justify that \( \log_b b = 1 \) for \( b > 0 \) and \( b \neq 1 \), we can fill in the blanks as follows:

Any positive number raised to the power 1 is equal to itself. Written as an equation, this means that \( b^1 = b \) for any positive base. This equation is equivalent to the logarithmic equation \( \log_b b = 1 \). Therefore, it must be true that \( \log_b b = 1 \) for any positive base that is not equal to 1.