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2^x^2=32(2^4x)
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\[2^{x^2}=32(2^{4x})\] \[2^{x^2}=2^{5(2^{4x})}\] \[x^2=5(2^{4x})\]
and then :) use the log ?
wait typo there
ohhhhhhhh
\[2^{x^2}=32(2^{4x})\] \[2^{x^2}=2^{5}{(2^{4x})}\] \[2^{x^2}=2^{5+4x}\] \[x^2=5+4x\] solve
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lol.. so easy :)) but I can't do it :) thanks a lot <3<3
yw
2x . 3^x= 7x . 5^x
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