Question

What is the solution of log2x + 727 = 3?

Answer 1

$log _{2x+7} 27 = 3$
( 2 x + 7 )³ = 27
( 2 x + 7 )³ = 3³
2 x + 7 = 3
2 x = 3 - 7
2 x = - 4
x = ( - 4 ) : 2
x = - 2

Answer 2

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.