What is the answer to 12(2)^x=15(4)^2x-3 ?

The answer to 12(2)^x=15(4)^2x-3 is

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12(2)^x=15(4)^2x-3

12 (2)^{x} = 15 (4)^{2} x - 3

x = - \frac{80 W _{- 1} ( - \frac{e ^{\frac{l n ( 2 )}{80}} l n ( 2 )}{20} ) - ln ( 2 )}{80 ln ( 2 )}, x = - \frac{80 W _{0} ( - \frac{e ^{\frac{l n ( 2 )}{80}} l n ( 2 )}{20} ) - ln ( 2 )}{80 ln ( 2 )}

Decimal Notation

x = 7.15920 \dots, x = 0.06479 \dots

Solution steps

12 (2)^{x} = 15 (4)^{2} x - 3

Prepare

12 (2)^{x} = 15 (4)^{2} x - 3

for Lambert form:

12 = e^{- l n (2) x} (240 x - 3)

Rewrite the equation with

(- x + \frac{1}{80}) ln (2) = u

and

x = - \frac{80 u - ln ( 2 )}{80 ln ( 2 )}

12 = e^{- l n (2) (- \frac{80 u - l n ( 2 )}{80 l n ( 2 )})} (240 (- \frac{80 u - ln ( 2 )}{80 ln ( 2 )}) - 3)

Simplify

12 = e^{\frac{80 u - l n ( 2 )}{80}} (- \frac{3 ( 80 u - ln ( 2 ) )}{ln ( 2 )} - 3)

Rewrite

12 = e^{\frac{80 u - l n ( 2 )}{80}} (- \frac{3 ( 80 u - ln ( 2 ) )}{ln ( 2 )} - 3)

in Lambert form:

e^{u} u = - \frac{e ^{\frac{l n ( 2 )}{80}} ln ( 2 )}{20}

Solve

e^{u} u = - \frac{e ^{\frac{l n ( 2 )}{80}} ln ( 2 )}{20} : u = W_{- 1} (- \frac{e ^{\frac{l n ( 2 )}{80}} ln ( 2 )}{20}), u = W_{0} (- \frac{e ^{\frac{l n ( 2 )}{80}} ln ( 2 )}{20})

u = W_{- 1} (- \frac{e ^{\frac{l n ( 2 )}{80}} ln ( 2 )}{20}), u = W_{0} (- \frac{e ^{\frac{l n ( 2 )}{80}} ln ( 2 )}{20})

Substitute back

u = (- x + \frac{1}{80}) ln (2),

solve for

x

Solve

(- x + \frac{1}{80}) ln (2) = W_{- 1} (- \frac{e ^{\frac{l n ( 2 )}{80}} ln ( 2 )}{20}) : x = - \frac{80 W _{- 1} ( - \frac{e ^{\frac{l n ( 2 )}{80}} l n ( 2 )}{20} ) - ln ( 2 )}{80 ln ( 2 )}

Solve

(- x + \frac{1}{80}) ln (2) = W_{0} (- \frac{e ^{\frac{l n ( 2 )}{80}} ln ( 2 )}{20}) : x = - \frac{80 W _{0} ( - \frac{e ^{\frac{l n ( 2 )}{80}} l n ( 2 )}{20} ) - ln ( 2 )}{80 ln ( 2 )}

x = - \frac{80 W _{- 1} ( - \frac{e ^{\frac{l n ( 2 )}{80}} l n ( 2 )}{20} ) - ln ( 2 )}{80 ln ( 2 )}, x = - \frac{80 W _{0} ( - \frac{e ^{\frac{l n ( 2 )}{80}} l n ( 2 )}{20} ) - ln ( 2 )}{80 ln ( 2 )}

4^{x + 3} = 7

(6 4^{x^{2} - 2 x}) (1 6^{x - 5}) = 1

(x + 2) e^{x} = 1

t, a (t) = a_{0} \cdot e^{- 0.025 t}

3 (4^{x}) + 4^{x - 1} = 52