Answer:
The solution of \(log_{2x+7}27\:=\:3\) is -2.
Step-by-step explanation:
Given : \(log_{2x+7}27\:=\:3\)
We have to solve for x.
Consider the given expression \(log_{2x+7}27\:=\:3\)
Apply rule, \(\log _a\left(b\right)=\frac{\ln \left(b\right)}{\ln \left(a\right)}\)
\(\log _{2x+7}\left(27\right)=\frac{\ln \left(27\right)}{\ln \left(2x+7\right)}\)
On simplifying, we get,
\(\frac{\ln \left(27\right)}{\ln \left(2x+7\right)}=3\)
Multiply both side by \(\ln \left(2x+7\right)\)
We have,
\(\frac{\ln \left(27\right)}{\ln \left(2x+7\right)}\ln \left(2x+7\right)=3\ln \left(2x+7\right)\)
Multiply fractions as \(\:a\cdot \frac{b}{c}=\frac{a\:\cdot \:b}{c}\)
we get,
\(=\frac{\ln \left(27\right)\ln \left(2x+7\right)}{\ln \left(2x+7\right)}==\ln \left(27\right)\)
Thus, becomes \(\ln \left(27\right)=3\ln \left(2x+7\right)\)
Divide both sides by 3, we have,
\(\frac{3\ln \left(2x+7\right)}{3}=\frac{\ln \left(27\right)}{3}\)
Apply rule \(\log _a\left(x^b\right)=b\cdot \log _a\left(x\right)\)
\(\ln(27)=\ln \left(3^3\right)=3\ln \left(3\right)\)
Thus, \(\ln \left(2x+7\right)=\ln \left(3\right)\)
\(\mathrm{When\:the\:logs\:have\:the\:same\:base:\:\:}\log _b\left(f\left(x\right)\right)=\log _b\left(g\left(x\right)\right)\quad \Rightarrow \quad f\left(x\right)=g\left(x\right)\)
2x+ 7 = 3
Subtract 7 both sides, we get,
2x = 3 - 7
Simplify , we have,
2x = -4
Divide both side by 2, we have
x = -2
Thus, the solution of \(log_{2x+7}27\:=\:3\) is -2.
