Evaluate :
log 4+13 log 125−15 log 32\text{log 4} + \dfrac{1}{3}\text{ log 125} - \dfrac{1}{5}\text{ log 32}log 4+31 log 125−51 log 32
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Evaluating the expression,
⇒log 4+13 log 125−15 log 32⇒log 4+13 log 53−15 log 25⇒log 4 + log (5)(3×13)−log (2)(5×15)⇒log 4 + log 5 - log 2⇒log 4×52⇒log 202⇒log 10⇒1.\Rightarrow \text{log 4} + \dfrac{1}{3}\text{ log 125} - \dfrac{1}{5}\text{ log 32} \\[1em] \Rightarrow \text{log 4} + \dfrac{1}{3}\text{ log 5}^3 - \dfrac{1}{5}\text{ log 2}^5 \\[1em] \Rightarrow \text{log 4 + log (5)}^{\Big(3 \times \dfrac{1}{3}\Big)} - \text{log (2)}^{\Big(5 \times \dfrac{1}{5}\Big)} \\[1em] \Rightarrow \text{log 4 + log 5 - log 2} \\[1em] \Rightarrow \text{log } \dfrac{4 \times 5}{2} \\[1em] \Rightarrow \text{log } \dfrac{20}{2} \\[1em] \Rightarrow \text{log } 10 \\[1em] \Rightarrow 1.⇒log 4+31 log 125−51 log 32⇒log 4+31 log 53−51 log 25⇒log 4 + log (5)(3×31)−log (2)(5×51)⇒log 4 + log 5 - log 2⇒log 24×5⇒log 220⇒log 10⇒1.
Hence, log 4+13 log 125−15 log 32\text{log 4} + \dfrac{1}{3}\text{ log 125} - \dfrac{1}{5}\text{ log 32}log 4+31 log 125−51 log 32 = 1.
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log 5 + log 8 - 2 log 2
log10 8 + log10 25 + 2 log10 3 - log10 18
Prove that :
2 log 1518−log 25162+log 49=log 2\text{2 log }\dfrac{15}{18} - \text{log } \dfrac{25}{162} + \text{log } \dfrac{4}{9} = \text{log 2}2 log 1815−log 16225+log 94=log 2
Find x, if :
x - log 48 + 3 log 2 = 13\dfrac{1}{3}31 log 125 - log 3