Question

solve log(10-4x)=log(10-3)

Answer 1

Log(10-3)? or is it Log(10-3x)?

Answer 2

\[\large \log(10-4x)=\log(10-3)\]
\[\large \log(10-4x)-\log(7)=0\]
Using your logarithm laws:
\[\large \log(a)-\log(b)=\log(\frac{a}{b})\]
\[\large \log(10-4x)-\log(7)=\log(\frac{10-4x}{7})\]
\[\large \log(\frac{10-4x}{7})=0\]
Using logarithm laws:
\[\log(a)=0\]
Can be written as:
\[10^0=a\]
\[a=1\]
Therefore applying this law to your question:
\[\large \frac{10-4x}{7}=10^0\]
\[\large \frac{10-4x}{7}=1\]
\[\large 10-4x=7\]
\[\large 4x=3\]
\[\large x=\frac{3}{4}\]

Answer 3

Hi thank you so much! and yeah it was log(10-3x)

Answer 4

Then that's the not the correct answer. Because I assumed you typed it correctly. Don't copy what I wrote!! You should really check what you write on the internet to see whether you made a typo of some sort. Then people would be able to show you how to solve your problem without giving you an incorrect result.

Answer 5

Here's the correct worked out solution:
\[\log(10-4x)=\log(10-3x)\]
\[\log(10-4x)-\log(10-3x)=0\]
\[\log(\frac{10-4x}{10-3x})=0\]
\[\frac{10-4x}{10-3x}=10^0\]
\[\frac{10-4x}{10-3x}=1\]
\[10-4x=10-3x\]
\[0=x\]
Therefore:
\[\huge x=0\]