Solve for x :
log 128log 32\dfrac{\text{log 128}}{\text{log 32}}log 32log 128 = x
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Given,
⇒log 128log 32=x⇒log 27log 25=x⇒7 log 25 log 2=x⇒x=75=1.4\Rightarrow \dfrac{\text{log 128}}{\text{log 32}} = x \\[1em] \Rightarrow \dfrac{\text{log 2}^7}{\text{log 2}^5} = x \\[1em] \Rightarrow \dfrac{\text{7 log 2}}{\text{5 log 2}} = x \\[1em] \Rightarrow x = \dfrac{7}{5} = 1.4⇒log 32log 128=x⇒log 25log 27=x⇒5 log 27 log 2=x⇒x=57=1.4
Hence, x = 1.4
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log (x + 5) + log (x - 5) = 4 log 2 + 2 log 3
log 81log 27\dfrac{\text{log 81}}{\text{log 27}}log 27log 81 = x
log 64log 8\dfrac{\text{log 64}}{\text{log 8}}log 8log 64 = log x
log 225log 15\dfrac{\text{log 225}}{\text{log 15}}log 15log 225 = log x
