# Logarithm Questions

Solution : $$2log{2\over 5}$$ + $$3log{25\over 8}$$ + $$log{128\over 625}$$ = $$log{2^2\over 5^2}$$ + $$log({5^2\over 2^3})^3$$ + $$log{2^7\over 5^4}$$ = $$log({2^2\over 5^2}{5^6\over 2^9}{2^7\over 5^4})$$ = log 1 = 0 Similar Questions Solve for x : $$2^{x + 2}$$ > $$({1\over 4})^{1\over x}$$. Evaluate the given log : $$81^{l\over {log_5 3}}$$ + $$27^{log_9 36}$$ + $$3^{4\over … ## Solve for x : \(2^{x + 2}$$ > $$({1\over 4})^{1\over x}$$.

Solution : We have $$2^{x + 2}$$ > $$2^{2/x}$$. Since the base 2 > 1, we have x + 2 > -$$2\over x$$ (the sign of the inequality is retained) Now x + 2 + $$2\over x$$  $$\implies$$ $$x^2 + 2x + 2\over x$$ > 0 $$\implies$$ $$(x + 1)^2 + 1\over x$$ > 0  …

## Evaluate the given log : $$81^{l\over {log_5 3}}$$ + $$27^{log_9 36}$$ + $$3^{4\over {log_7 9}}$$.

Solution : $$81^{log_3 5}$$ + $$3^{3log_9 36}$$ + $$3^{4log_9 7}$$ $$\implies$$ $$3^{4log_3 5}$$ + $$3^{log_3 {(36)}^{3/2}}$$ + $$3^{log_3 {7}^2}$$ = 625 + 216 + 49 = 890. Similar Questions Solve for x : $$2^{x + 2}$$ > $$({1\over 4})^{1\over x}$$. Find the value of $$2log{2\over 5}$$ + $$3log{25\over 8}$$ – $$log{625\over 128}$$. If $$log_a x$$ …

## If $$log_a x$$ = p and $$log_b {x^2}$$ = q then $$log_x \sqrt{ab}$$ is equal to

Solution : $$log_a x$$ = p $$\implies$$ $$a^p$$ = x $$\implies$$ a = $$x^{1/p}$$ Similarly  $$b^q$$ = $$x^2$$ $$\implies$$ b = $$x^{2/q}$$ Now, $$log_x \sqrt{ab}$$ = $$log_x \sqrt{x^{1/p}x^{2/q}}$$ = $$log_x x^{({1\over p}+{2\over q}){1\over 2}}$$ = $$1\over {2p}$$ + $$1\over q$$. Similar Questions Solve for x : $$2^{x + 2}$$ > $$({1\over 4})^{1\over x}$$. Evaluate the …

## If $$log_e x$$ – $$log_e y$$ = a, $$log_e y$$ – $$log_e z$$ = b & $$log_e z$$ – $$log_e x$$ = c, then find the value of $$({x\over y})^{b-c}$$ $$\times$$ $$({y\over z})^{c-a}$$ $$\times$$ $$({z\over x})^{a-b}$$.

Solution : $$log_e x$$ – $$log_e y$$ = a $$\implies$$ $$log_e {x\over y}$$ = a $$\implies$$ $$x\over y$$ = $$e^a$$ $$log_e y$$ – $$log_e z$$ = b $$\implies$$ $$log_e {y\over z}$$ = b $$\implies$$ $$y\over z$$ = $$e^b$$ $$log_e z$$ – $$log_e x$$ = c $$\implies$$ $$log_e {z\over x}$$ = c $$\implies$$ $$z\over x$$ = …