# How do you solve logy = log16 + log49?

Jul 6, 2018

I tried this:

#### Explanation:

I would first use a property of logs to write:

$\log y = \log \left(16 \cdot 49\right)$

then I would use the definition of log

${\log}_{b} x = a$
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:

$y = 16 \cdot 49 = 784$

Jul 6, 2018

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