Porisma

Worked solution

Solve an equation with three exponential bases

Solve the equation 3x+1⋅4x+2=5x+33^{x+1}\cdot4^{x+2}=5^{x+3}.

CJC · Discussion · Question 11

Answer

x=ln⁡(125/48)/ln⁡(12/5)≈1.09x=\ln(125/48)/\ln(12/5)\approx1.09.

Take logarithms and collect x

(x+1)ln⁡3+(x+2)ln⁡4=(x+3)ln⁡5(x+1)\ln3+(x+2)\ln4=(x+3)\ln5. Thus x(ln⁡3+ln⁡4−ln⁡5)=3ln⁡5−2ln⁡4−ln⁡3x(\ln3+\ln4-\ln5)=3\ln5-2\ln4-\ln3, giving x=3ln⁡5−2ln⁡4−ln⁡3ln⁡3+ln⁡4−ln⁡5≈1.09x=\dfrac{3\ln5-2\ln4-\ln3}{\ln3+\ln4-\ln5}\approx1.09.

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Alternatively combine the exponential factors

3x4x5x=533⋅42\dfrac{3^x4^x}{5^x}=\dfrac{5^3}{3\cdot4^2}, so (12/5)x=125/48(12/5)^x=125/48. Hence x=ln⁡(125/48)ln⁡(12/5)≈1.09x=\dfrac{\ln(125/48)}{\ln(12/5)}\approx1.09 (3 significant figures).

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Common mistakes

Apply logarithms to products by adding the logarithms, not multiplying them.

Read the hint