Express Log103 + 1 in Terms of Log10x. - Mathematics

Express log₁₀3 + 1 in terms of log₁₀x.

बेरीज

log₁₀3 + 1
= log₁₀ 3 + log₁₀ 10
= log₁₀ (3 x 10)
= log₁₀30.

shaalaa.com

More About Logarithm

या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?

पाठ 10: Logarithms - Exercise 10.2

पाठ 10 Logarithms
Exercise 10.2 | Q 9

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.

If log_√27x = 2 `(2)/(3)` , find x.

Find x, if : log_x 625 = - 4

If log₂(x + y) = log₃(x - y) = `log 25/log 0.2`, find the values of x and y.

Solve for x, `log_x^(15√5) = 2 - log_x^(3√5)`.

Solve the following:
log (3 - x) - log (x - 3) = 1

Solve the following:
log ( x + 1) + log ( x - 1) = log 11 + 2 log 3

Solve for x: `("log"1331)/("log"11)` = logx

If `"log" x^2 - "log"sqrt(y)` = 1, express y in terms of x. Hence find y when x = 2.

Prove that: `(1)/("log"_2 30) + (1)/("log"_3 30) + (1)/("log"_5 30)` = 1