Asked by Shellie
Evaluate log b {square root of 10b}, given that log b 2 = 0.3562 and
log b 5 = 0.8271 (sorry I had no clue how to type this in math format!)
log b 5 = 0.8271 (sorry I had no clue how to type this in math format!)
Answers
Answered by
MathMate
I am not sure if I interpret your question correctly.
Given log<sub>b</sub>2 = 0.3562 and
log<sub>b</sub>5 = 0.8271,
evaluate log<sub>b</sub>{square-root of 10<sub>b</sub>};
log<sub>b</sub>{square-root of 10<sub>b</sub>}
= log<sub>b</sub>{square-root of 2<sub>b</sub>*5<sub>b</sub>}
=(1/2)log<sub>b</sub>(10<sub>b</sub>)
=(1/2)(1)
=0.5
since log<sub>b</sub>10<sub>b</sub> = b for any b.
However, if the number 10 is replaced by 10 to the base 10, we proceed slightly differently:
log<sub>b</sub>{square-root of 10}
=(1/2)log<sub>b</sub>(10)
= (1/2)log<sub>b</sub>(2*5)
= (1/2)(log<sub>b</sub>(2)+log<sub>b</sub>(5))
= (1/2)(0.3562+0.8271)
=0.5916
Given log<sub>b</sub>2 = 0.3562 and
log<sub>b</sub>5 = 0.8271,
evaluate log<sub>b</sub>{square-root of 10<sub>b</sub>};
log<sub>b</sub>{square-root of 10<sub>b</sub>}
= log<sub>b</sub>{square-root of 2<sub>b</sub>*5<sub>b</sub>}
=(1/2)log<sub>b</sub>(10<sub>b</sub>)
=(1/2)(1)
=0.5
since log<sub>b</sub>10<sub>b</sub> = b for any b.
However, if the number 10 is replaced by 10 to the base 10, we proceed slightly differently:
log<sub>b</sub>{square-root of 10}
=(1/2)log<sub>b</sub>(10)
= (1/2)log<sub>b</sub>(2*5)
= (1/2)(log<sub>b</sub>(2)+log<sub>b</sub>(5))
= (1/2)(0.3562+0.8271)
=0.5916
Answered by
Reiny
I saw the question this way
log<sub>b</sub> (√(10b))
= 1/2(log<sub>b</sub>(10b))
= 1/2(log<sub>b</sub>5 + log<sub>b</sub>2 + log<sub>b</sub>b)
= 1/2(.8271 + .3562 + 1)
= 1.0917
log<sub>b</sub> (√(10b))
= 1/2(log<sub>b</sub>(10b))
= 1/2(log<sub>b</sub>5 + log<sub>b</sub>2 + log<sub>b</sub>b)
= 1/2(.8271 + .3562 + 1)
= 1.0917
Answered by
Shellie
The second answer is the correct answer! Thanks to you both I am on my way to the final
Shellie
Shellie
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