Write as a single log, please show working.

2log6 - 4(log3 + log2)

Here is what I got:

= log6^2 - log2^4 + log2^4

= log6^2 + log2^4 - log2^4

= log(6^2 + 2^4) - log3^4

= log52 - log 3^4

= log 52/81

2 posts
• Page **1** of **1**

Write as a single log, please show working.

2log6 - 4(log3 + log2)

Here is what I got:

= log6^2 - log2^4 + log2^4

= log6^2 + log2^4 - log2^4

= log(6^2 + 2^4) - log3^4

= log52 - log 3^4

= log 52/81

2log6 - 4(log3 + log2)

Here is what I got:

= log6^2 - log2^4 + log2^4

= log6^2 + log2^4 - log2^4

= log(6^2 + 2^4) - log3^4

= log52 - log 3^4

= log 52/81

- dgware
**Posts:**1**Joined:**Mon Mar 31, 2014 3:17 am

dgware wrote:Write as a single log, please show working.

2log6 - 4(log3 + log2)

Here is what I got:

= log6^2 - log2^4 + log2^4

No. The 1st log is of 3, not of 2. Plus -4(log3 + log2) = -4(log3) - 4(log2). So fix the signs.

- theshadow
**Posts:**91**Joined:**Sun Feb 22, 2009 11:12 pm

2 posts
• Page **1** of **1**