Question

log2(3x+1)=4

Answer 1

solve for x

Answer 2

use the log property that
\(\Large \log_ab=c \implies b=a^c\)

Answer 3

\(\bf log_2(3x+1)=4\\ \quad \\
\textit{recall the log cancellation rule of}\quad a^{log_ax} =x\qquad thus\\ \quad \\
log_2(3x+1)=4\implies 2^{log_2(3x+1)}=2^4\implies (3x+1)=2^4\)

Answer 4

I got 2^4 = 3x+1, 16=3x+1 and x = 5

Answer 5

can you also help me with log(x-3) + log(x)=1?

Answer 6

http://www.chilimath.com/algebra/advanced/log/images/466x428xrules,P20of,P20exponents.gif.pagespeed.ic.1z7-Ekl5L5.png <-- notice the 1st rule listed there

Answer 7

(x-3)(x) = 1 ? or (x-3)(x)=10 because of log?

Answer 8

well... ... yes

Answer 9

\(\bf \large {log(x-3) + log(x)=1\implies log[x(x-3)]=1\\ \quad \\
10^{log_{10}[x(x-3)]}=10^1\implies x(x-3)=10
}\)

Answer 10

got it, thanks so much!

Answer 11

yw