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4^4x - 1 = 8^2xSolve for x.

Sagot :

x = 1

Explanation:

[tex]4^{4x-1}=8^{2x}[/tex]

Make the base have the same number:

8 = 2³

4 = 2²

The base becomes 2

[tex]2^{2(4x-1)}=2^{3(}^{2x)}[/tex]

Then we simplify:

[tex]\begin{gathered} \text{The base cancels out:} \\ 2(4x-1)\text{ = 3(2x)} \\ 8x\text{ - 2 = 6x} \\ \text{collect like terms: } \\ 8x\text{ - 6x = 2} \\ 2x\text{ = 2} \\ x\text{ = 2/2} \\ x\text{ = 1} \end{gathered}[/tex]

Therefore, x = 1

check:

[tex]\begin{gathered} 4^{4(1)-1}=8^{2(1)} \\ 4^{4-1}=8^2 \\ 4^3\text{ }=8^2 \\ 4\times4\times4=\text{ 64} \\ 8\times8\text{ = 64} \\ 64\text{ = 64 (x=1 is correct)} \end{gathered}[/tex]