Asked by isa

solve exactly for a in the equation:

log a^5 + log a^(3/2) = log 625
4 2 8

the 4,2,8 are the bases
Answered by isa
log4 a^5 + log2 a^(3/2) = log8 625
Answered by Reiny
let's convert to the same base 2

log<sub>4</sub> a^5 = log<sub>2</sub> a^5 /log<sub>2</sub> 4
= (1/2)(log<sub>2</sub> a^5) since log<sub>2</sub> 4 = 2

similary log<sub>8</sub> 625
= log<sub>2</sub> 625/log<sub>2</sub> 8

so the new equation is
(1/2)(log<sub>2</sub> a^5) + log<sub>2</sub> a^(3/2) = (1/3)log<sub>2</sub>625

using the rules of logs

log<sub>2</sub> a^(5/2) + log<sub>2</sub> a^(3/2) = log<sub>2</sub> 625^(1/3)

log<sub>2</sub> a^4 = log<sub>2</sub> 625^(1/3)

a^4 = 625^(1/3)
a = 5^(1/3) or the cuberoot of 5

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