Question

2log5/8+log128/125+log5/2=0

Answer 1

Refer to the attachment

Answer 2

$LHS = 2log \: \frac{5}{8} + log \: \frac{128}{125} +log \: \frac{5}{2} \\= log \: \Big(\frac{5}{8}\Big)^{\pink {2}} + log \: \frac{128}{125} +log \: \frac{5}{2}$

$\boxed { \orange { nlog \: a = log \:a^{n} }}$

$= log \: \Big(\frac{5}{8}\Big)^ {2}\times \frac{128}{125} \times \frac{5}{2}$

$\boxed { \orange { log m + log n = log ( m \times n ) }}$

$= log \:\frac{ 5^{2} \times 128 \times 5 }{8^{2} \times 125 \times 2} \\= log \:\frac{ 5\times 5 \times 8\times 8 \times 2 \times 5 }{8\times 8\times 5 \times 5 \times 5 \times 2} \\= log \frac{2\times 5^{3} \times 8^{2}}{2\times 5^{3} \times 8^{2}} \\= log \:1\\= 0 \\= RHS$

Therefore.,

$\red {2log \: \frac{5}{8} + log \: \frac{128}{125} +log \: \frac{5}{2}} \green {= 0 }$

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