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Solve for x: 81^ЁЭСетИТ2 = 27^ЁЭСе+1 - Mathematics

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Solve for x:

`81^(x - 2) = 27^(x + 1)`

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Given expression is `81^(x - 2) = 27^(x + 1)`.

We have to find the value of x in given expression.

Thus, `81^(x - 2) = 27^(x + 1)`

`(3^4)^(x - 2) = (3^3)^(x + 1)`

`(3)^(4(x - 2)) = (3)^(3(x + 1))`   ...[∴ (an)m = anm]

`(3)^(4x - 8) = (3)^(3x + 3)`

Equating the powers with same bases.

4x – 8 = 3x + 3

4x – 3x = 3 + 8

x = 11

Therefore, the value of x in given expression is 11.

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рдЕрдзреНрдпрд╛рдп 6: Indices - EXERCISE 6 [рдкреГрд╖реНрда ремрен]

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рдмреА рдирд┐рд░реНрдорд▓рд╛ рд╢рд╛рд╕реНрддреНрд░реА Mathematics [English] Class 9 ICSE
рдЕрдзреНрдпрд╛рдп 6 Indices
EXERCISE 6 | Q 11. (ii) | рдкреГрд╖реНрда ремрен
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