Exercise 78 Page 971

To solve the given logarithmic equation, we will first recall the definition of a logarithm. log_b x=y ⇔ x= b^y This tells us how we can rewrite the logarithm equivalent to y as an exponential equation. The argument x is equal to b raised to the power of y. The base of a natural logarithm is e, so ln x = log_e x. To use this definition, we need to rewrite the given equation.

2ln 4x +5 = 8

SubEqn

LHS-5=RHS-5

2ln 4x = 3

b*ln(a)=ln(a^b)

ln (4x)^2 = 3

Now, let's rewrite this equation in exponential form. ln (4x)^2= 3 ⇔ (4x)^2= e^3 We will solve this equation for x. Let's do it!

(4x)^2=e^3

SqrtEqn

sqrt(LHS)=sqrt(RHS)

4x=± sqrt(e^3)

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Solve for x

WriteSum

Write as a sum

4x=± sqrt(e^(1+2))

SumInExponentOne

a^(1+m)=a*a^m

4x=± sqrt(e* e^2)

CommutativePropMult

Commutative Property of Multiplication

4x=± sqrt(e^2* e)

SqrtProd

sqrt(a* b)=sqrt(a)*sqrt(b)

4x=± sqrt(e^2)* sqrt(e)

sqrt(a^2)=± a

4x=± esqrt(e)

DivEqn

.LHS /4.=.RHS /4.

x=± esqrt(e)/4

The exact solutions are x_1= esqrt(e)4 and x_2= - esqrt(e)4. We can also write them in decimal form using a calculator. x_1&=&esqrt(e)/4 &≈ 1.120 x_2&=&- esqrt(e)/4 &≈ - 1.120