Question

solve for x : log2 3x=2=1
the first 2 is lower than the word log

log5(2x+1)=1
5 is lower than the word log

Answer 1

I don't know u write (3x+2) or (3x-2), If (3x+2) than x=0
\[\log _{2}^{(3x-2)}=1\]\[(3x-2)\log _{2}=\log(1)\]\[3x-2=(\log 1)(\log2)\]\[3x-2=2\]\[3x=2+2\]\[x=\frac{4}{3}\]and another\[\log _{5}(2x+1)=1\]\[(2x+1)\log _{5}=\log(1)\]\[2x+1=(\log(1)) \times(\log(5))\]\[2x+1=5\]\[2x=5-1\]\[x=\frac{4}{2}\]\[x=2\]

Answer 2

lets try this. for the second one rewrite in equivalent exponential form. saying
\[b^x=y\] same as
\[\log_b(y)=x\]

Answer 3

you have \[\log_5(2x+1)=1\] this means the same thing as
\[2x+1=5^1=5\] and now you solve in your head to get \[x=2\]

Answer 4

try as i might i cannot makes heads nor tails of the solution given above. first of all there is no such thing as
\[log_5\] without something after it. second there is no way
\[log_5(2x+1)=(2x+1)log_5\] this has not meaning at all.
third
\[log_5(1)=0\] because
\[b^0=1\] and of course
\[log_b(1)=0\] for all b.

Answer 5

so this line
\[2x+1=(log(1))×(log(5))\]
says
\[2x+1=0\]

Answer 6

just remember to write in equivalent exponential form:
\[log_b(x)=y \Leftrightarrow b^y=x\] and you should be able to switch back and forth with ease