Question

Log10 8 +2log10 2-log10 32+log10 5

Answer 1

Answer:

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Answer 2

$<b>$

Log10=1

$formulae : log {a}^{b} = bloga$

now,

$log {10}^{8} + 2log {10}^{2} - log {10}^{32} + log {10}^{5} \\ \\ 8 \times 1 + 2 \times 2 \times 1 - 32 \times 1 + 5 \times 1 \\ \\ 8 + 4 - 32 + 5 \\ \\ - 23$