Question

How do you solve `logy = log16 + log49`?

Answer 1

I tried this:

Explanation:

I would first use a property of logs to write:

`logy=log(16*49)`

then I would use the definition of log

`log_bx=a`
so that:
`x=b^a`
and the relationship between exponential and log:

considering that your logs are in base `b` (whatever `b` could be) to write:

`b^(log_by)=b^(log_b(16*49)`

giving:

`y=16*49=784`

Answer 2

The goal is to achieve an expresion like `log A=log B` and from unicity of logarithm we will get `A=B`

In our case `logy=log16+log49=log(16·49)`, then `y=16·49=784`

We have used `log(x·y)=logx+logy`