![If log(0.5), log(2x – 1) and log(2x + 5) are in arithmetic progression, then the value of x is - YouTube If log(0.5), log(2x – 1) and log(2x + 5) are in arithmetic progression, then the value of x is - YouTube](https://i.ytimg.com/vi/if0ceXAKBLU/hqdefault.jpg)
If log(0.5), log(2x – 1) and log(2x + 5) are in arithmetic progression, then the value of x is - YouTube
![Limits Ex 29.10021 \( \lim _ { \log ( 2 + x ) + \log ( 0.5 } \) \( = 1 \mathrm { im } \) \( = \frac { 1 } { 2 } \) \( \frac { \log \left( 1 + \frac { x } { 2 } \right) } { 2 \left( \frac { x } { 2 } \right) } \) Limits Ex 29.10021 \( \lim _ { \log ( 2 + x ) + \log ( 0.5 } \) \( = 1 \mathrm { im } \) \( = \frac { 1 } { 2 } \) \( \frac { \log \left( 1 + \frac { x } { 2 } \right) } { 2 \left( \frac { x } { 2 } \right) } \)](https://toppr-doubts-media.s3.amazonaws.com/images/2092981/3daa5da1-23ed-4c31-80af-f3be408aac0f.jpg)
Limits Ex 29.10021 \( \lim _ { \log ( 2 + x ) + \log ( 0.5 } \) \( = 1 \mathrm { im } \) \( = \frac { 1 } { 2 } \) \( \frac { \log \left( 1 + \frac { x } { 2 } \right) } { 2 \left( \frac { x } { 2 } \right) } \)
![Log (base 0 5x/10)of x^2 - 14log (base 16x)of x^3 + 40log (base 4x)of x^(1/2) solve for x - Maths - Linear Inequalities - 14461313 | Meritnation.com Log (base 0 5x/10)of x^2 - 14log (base 16x)of x^3 + 40log (base 4x)of x^(1/2) solve for x - Maths - Linear Inequalities - 14461313 | Meritnation.com](https://s3mn.mnimgs.com/img/shared/content_ck_images/ck_5ea6cf91be624.jpg)
Log (base 0 5x/10)of x^2 - 14log (base 16x)of x^3 + 40log (base 4x)of x^(1/2) solve for x - Maths - Linear Inequalities - 14461313 | Meritnation.com
![2013 - 2012 3—2 Us x : 1024 — -0.4251 3) log2x 2) 3 ) log 0.5 + log 2 4) X (32) 3 5) 210g2 6 - [PDF Document] 2013 - 2012 3—2 Us x : 1024 — -0.4251 3) log2x 2) 3 ) log 0.5 + log 2 4) X (32) 3 5) 210g2 6 - [PDF Document]](https://demo.documents.pub/img/378x509/reader022/reader/2020052003/5e48c89ebea89d3ded33ffd1/r-1.jpg)