O lvl A Maths: Indices and Logarithms

(a) Solve for x in the equation
(2x)^lg2 = (7x)^lg7

(b) Solve (log_aX)^log_bX = X where a, b are positive real numbers except 1, leave the answer in terms of a and b.

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Answer:

(a)
(2x)^{lg 2} = (7x)^lg7
(2^{lg 2})(x^{lg 2})=(7^{lg 7})(x^{lg 7})
(x^{lg 2}) / (x^{lg 7})=(7^{lg 7}) / (2^{lg 2})
x^{(lg2 - lg7)} = (7^{lg 7}) / (2^{lg 2})

Bring over the power to the other side of the equation
x = [(7^{lg 7}) /(2^{lg 2})]^{{1/(lg2-lg7)}}


(b)
(log_a x)^{log_b x} = x

Let x = b^{log_b x}
(loga x)^{log_b x} = b^{log_b x}

Same power, equate base
log_a x = b
x = a^b

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