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how do i solve this log2 3+log2 x=3
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then log2(3x)=3 2^3=3x 8=3x x=8/3
ok great :)thats what i got
i need help with another one
well using log laws log(ab) = log(a) + log(b) \[\log_{2}(3 \times x) = 3\] which when simplified is \[\log_{2}(3x) = 3\] so now raise each side as a power of the base \[2^{\log_{2}(3x)} = 2^3\] which simplifies to 3x = 8 now just solve for x
ok just write it
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3^2x+2=8 solve for x..
i got -0.15
3^2x=6 ln3^2x=ln6 2x ln3=ln6 2x=ln6/ln3 2x=1.630929754 x=0.815464876
the exponent is 3^(2x+2)
oh, i thought the +2 was out of exponent wait a second
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thanks :)
2x+2=ln8/ln3 2x+2=1.892789261 x=-0.053605369
yay perfect :D
thank you :)
you're welcome :)
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