X- 0lim loga 1-x x-logae- a-0- - 1

We have, #10; #10; x→ 0lim log #10;_a( 1+x )x=log_ae We know that x→ 0lim log_a( 1+x )x=1 #10; #10;Therefore, #10; #10; log_ae=1 ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Mathematics

Global Maxima

x→ 0lim loga ...

Question

${lim}_{x \to 0} \frac{{log}_{a} (1 + x)}{x} = {log}_{a} e, a > 0, \neq 1$

Open in App

Solution

We have,

${lim}_{x \to 0} \frac{{log}_{a} (1 + x)}{x} = {log}_{a} e$

We know that
${lim}_{x \to 0} \frac{{log}_{a} (1 + x)}{x} = 1$

Therefore,

${log}_{a} e = 1$
${log}_{a} e = l o g_{a} a$

Therefore,
$a = e$

Hence, this is the answer.

Suggest Corrections

Similar questions

Q. Let

a > 0, a \neq 1

and

x > 0

(i)

{log}_{a} x > 0, a > 1 \Leftrightarrow x > 0, a > 1

(ii)

{log}_{a} x > 0, 0 < a < 1 \Leftrightarrow 0 < x < 1, 0 < a < 1

(iii)

{log}_{a} x > 0, a > 1 \Leftrightarrow 0 < x < 1, a > 1

(iv)

{log}_{a} x < 0, 0 < a < 1 \Leftrightarrow x > 1, 0 < a < 1

What is the solution set of

{log}_{3} \frac{5 - 2 x}{12 x - 8} + {log}_{1 / 3} x \leq 0

Q. Based on this information answer the given queston:
Let

a > 0, a \neq 1

and

x > 0

(i)

{log}_{a} x > 0, a > 1 \Leftrightarrow x > 0, a > 1

(ii)

{log}_{a} x > 0, 0 < a < 1 \Leftrightarrow 0 < x < 1, 0 < a < 1

(iii)

{log}_{a} x > 0, a > 1 \Leftrightarrow 0 < x < 1, a > 1

(iv)

{log}_{a} x < 0, 0 < a < 1 \Leftrightarrow x > 1, 0 < a < 1

Identify the solution set of

x

{log}_{1 / 10} (x^{2} + x + 1) > 0

Q. Based on this information answer the questions given
If

(a > 1, x > 1)

(0 < a < 1, 0 < x < 1)

.
then

{log}_{a} x > 0

i.e.,

{log}_{a} x

is positive.
If

(0 < a < 1, x > 1)

(a > 1, 0 < x < 1)

, then

{log}_{a} x < 0

a > b

then

{log}_{b} a > 1

and

{log}_{a} b < 1.

Determine the sign of

\frac{{log}_{a} (3.162)}{{log}_{a} (2 / 3)} .

Q. Assertion :If

{log}_{0.2} (\frac{x + 1}{x}) \geq 1

, then

x \in [- 1.25, - 1)

. Reason: If

0 < a < 1

{log}_{a} x \geq {log}_{a} y \Leftrightarrow y \geq x > 0

Based on this information answer the question given

(a > 1, x > 1)

(0 < a < 1, 0 < x < 1)

then

{log}_{a} x > 0

i.e.,

{log}_{a} x

is positive.

(0 < a < 1, x > 1)

(a > 1, 0 < x < 1)

, then

{log}_{a} x < 0

a > b

then

{log}_{b} a > 1

and

{log}_{a} b < 1

Which is greater

{log}_{2} 11

{log}_{11} 2

Join BYJU'S Learning Program

limx→0loga(1+x)x=logae,a>0,≠1

We have, limx→0loga(1+x)x=logaeWe know thatlimx→0loga(1+x)x=1 Therefore, logae=1logae=logaaTherefore,a=e Hence, this is the answer.

${lim}_{x \to 0} \frac{{log}_{a} (1 + x)}{x} = {log}_{a} e, a > 0, \neq 1$

We have,

${lim}_{x \to 0} \frac{{log}_{a} (1 + x)}{x} = {log}_{a} e$

We know that
${lim}_{x \to 0} \frac{{log}_{a} (1 + x)}{x} = 1$

Therefore,

${log}_{a} e = 1$
${log}_{a} e = l o g_{a} a$

Therefore,
$a = e$

Hence, this is the answer.