Question

solving equations with fractional exponents
score: 0/2 penalty: none
question
find the positive solution of the equation.
$5x^{\frac{4}{3}} - 25 = 11980$

Explanation:

Step1: Isolate the term with the fractional exponent

Add 25 to both sides of the equation \(5x^{\frac{1}{4}} - 25 = 11980\) to get \(5x^{\frac{1}{4}}=11980 + 25\).
Simplifying the right side: \(5x^{\frac{1}{4}}=12005\)

Step2: Solve for \(x^{\frac{1}{4}}\)

Divide both sides of the equation \(5x^{\frac{1}{4}} = 12005\) by 5.
We have \(x^{\frac{1}{4}}=\frac{12005}{5}\)
Simplifying the right side: \(x^{\frac{1}{4}} = 2401\)

Step3: Solve for \(x\)

Raise both sides of the equation \(x^{\frac{1}{4}}=2401\) to the power of 4 to eliminate the fractional exponent.
Using the property \((a^m)^n=a^{mn}\), we get \(x=(2401)^4\)
But we know that \(2401 = 7^4\) (since \(7\times7\times7\times7 = 2401\)), so \(x=(7^4)^4\)
Using the property \((a^m)^n=a^{mn}\) again, \(x = 7^{16}\)? Wait, no, wait. Wait, actually, if \(x^{\frac{1}{4}}=2401\), and \(2401=7^4\), then \(x^{\frac{1}{4}}=7^4\). Raising both sides to the 4th power: \(x=(7^4)^4=7^{16}\)? Wait, no, that's a mistake. Wait, \(x^{\frac{1}{4}}=2401\), so to solve for \(x\), we can write \(x=(2401)^4\), but 2401 is \(7^4\), so \((7^4)^4=7^{16}\)? Wait, no, wait, \(x^{\frac{1}{4}} = 2401\), so \(x=(2401)^4\). But 2401 is \(7^4\), so \((7^4)^4=7^{16}\)? Wait, no, that's incorrect. Wait, let's check again. Wait, \(x^{\frac{1}{4}}=2401\), so \(x=(2401)^4\). But 2401 is \(7^4\), so \((7^4)^4 = 7^{16}\)? Wait, no, \((a^m)^n=a^{m\times n}\), so \(m = 4\), \(n = 4\), so \(4\times4 = 16\), so \(7^{16}\)? Wait, but 2401 is \(7^4\), so \(x^{\frac{1}{4}}=7^4\), then raising both sides to the 4th power: \(x=(7^4)^4=7^{16}\)? Wait, that seems too big. Wait, maybe I made a mistake in the value of 2401. Wait, \(7\times7 = 49\), \(49\times7=343\), \(343\times7 = 2401\). Yes, \(7^4=2401\). So \(x^{\frac{1}{4}}=7^4\), then \(x=(7^4)^4=7^{16}\)? Wait, no, that's not right. Wait, the equation is \(x^{\frac{1}{4}}=2401\), so \(x = 2401^4\). But 2401 is \(7^4\), so \(x=(7^4)^4=7^{16}\). But let's check with a smaller approach. Wait, maybe I made a mistake in step 2. Wait, \(5x^{\frac{1}{4}}-25 = 11980\), adding 25: \(5x^{\frac{1}{4}}=12005\), dividing by 5: \(x^{\frac{1}{4}}=2401\). Then, since \(x^{\frac{1}{4}}=\sqrt[4]{x}\), so to solve for \(x\), we take both sides to the 4th power: \(x=(2401)^4\). But 2401 is \(7^4\), so \(x=(7^4)^4 = 7^{16}\). But let's compute \(7^4=2401\), \(7^8=(7^4)^2=2401^2 = 5764801\), \(7^{12}=7^8\times7^4=5764801\times2401\), \(7^{16}=7^{12}\times7^4\). But that seems very large. Wait, maybe I made a mistake in the problem. Wait, the original equation is \(5x^{\frac{1}{4}}-25 = 11980\). Let's re - check the arithmetic. 11980+25 = 12005. 12005 divided by 5 is 2401. Then \(x^{\frac{1}{4}} = 2401\), so \(x = 2401^4\). But 2401 is \(7^4\), so \(x=(7^4)^4=7^{16}\). Alternatively, we can note that \(2401 = 7^4\), so \(x^{\frac{1}{4}}=7^4\), then \(x=(7^4)^4=7^{16}\). But let's verify: If \(x = 7^{16}\), then \(x^{\frac{1}{4}}=(7^{16})^{\frac{1}{4}}=7^4 = 2401\). Then \(5x^{\frac{1}{4}}=5\times2401 = 12005\), and \(12005-25 = 11980\), which matches the original equation. So the solution is correct.

Answer:

\(x = 7^{16}\) or \(x=33232930569601\) (if we calculate \(7^{16}\): \(7^2 = 49\), \(7^4=49^2 = 2401\), \(7^8 = 2401^2=5764801\), \(7^{16}=5764801^2 = 33232930569601\))