How do you solve #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:

#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 · 2018-07-06T12:35:54

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#