How do you solve $log y = log 16 + log 49$ $logy = log16 + log49$ ?

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Answer 1 · 2018-07-06T12:35:32

I tried this:

I would first use a property of logs to write:

$log y = log (16 \cdot 49)$ $logy=log(16*49)$

then I would use the definition of log

${log}_{b} x = a$ $log_bx=a$
so that:
$x = b^{a}$ $x=b^a$
and the relationship between exponential and log:

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

$b^{{log}_{b} y} = b^{{log}_{b} (16 \cdot 49)}$ $b^(log_by)=b^(log_b(16*49)$

giving:

$y = 16 \cdot 49 = 784$ $y=16*49=784$

Answer 2 · 2018-07-06T12:35:54

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

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

We have used $log (x \cdot y) = log x + log y$ $log(x·y)=logx+logy$

How do you solve logy=log16+log49logy = log16 + log49?