Solve for x: log⁡125log⁡5 = logx

Solve for x: `("log"125)/("log"5)` = logx

बेरीज

`("log"125)/("log"5)` = logx

⇒ `("log"5^3)/("log"5)` = logx

⇒ `(3"log"5)/("log"5)` = logx
⇒ 3 = logx
⇒ 3log10 = log x ...(since log 10 = 1)
⇒ log 10³ = logx
∴ x = 10³
= 1000

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More About Logarithm

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पाठ 10: Logarithms - Exercise 10.2

पाठ 10 Logarithms
Exercise 10.2 | Q 8.5

If x = 1 + log 2 - log 5, y = 2 log3 and z = log a - log 5; find the value of a if x + y = 2z.

Evaluate :`1/( log_a bc + 1) + 1/(log_b ca + 1) + 1/ ( log_c ab + 1 )`

If x = log 0.6; y = log 1.25 and z = log 3 - 2 log 2, find the values of :
(i) x+y- z
(ii) 5^{x + y - z}

If log`( a - b )/2 = 1/2( log a + log b )`, Show that : a² + b² = 6ab.

Given x = log₁₀12 , y = log₄ 2 x log₁₀9 and z = log₁₀0.4 , find :
(i) x - y - z
(ii) 13^{x - y - z}

If log ₃ m = x and log₃ n = y, write down
`3^(1-2y+3x)` in terms of m an n

Express the following in a form free from logarithm:
`2"log" x + 1/2"log" y` = 1

Prove that (log a)² - (log b)² = `"log"("a"/"b")."log"("ab")`

If a b + b log a - 1 = 0, then prove that b^a.a^b = 10

Prove that log ₁₀125 = 3 (1 - log ₁₀ 2)