If $$log52$$,$$log52x$$−$$3$$,$$log5172+2x-1$$⁡ are in A.P then x is

$$2log52x$$−$$3=log52$$ $$+log5172+2x-1$$⇒$$2x$$−$$32=17+2x$$ Let $$2x=y$$⇒$$y2$$−$$7y$$−$$8=0$$→$$y=8$$ or $$y=-1$$, which is not possible since y>$$0$$ If $$y=8$$→$$2x=8x=3$$.

Arithmetic progression

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$If \log_{5}^{2}, \log_{5}^{(2^{x} - 3)}, \log_{5}^{(\frac{17}{2} + 2^{x - 1})} are in A.P then x is$

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Solution

$2 \log_{5}^{(2^{x} - 3)} =$ $\log_{5}^{2} + \log_{5}^{(\frac{17}{2} + 2^{x - 1})}$
$\begin{array}{l} \Rightarrow {(2^{x} - 3)}^{2} = (17 + 2^{x}) \\ Let 2^{x} = y \\ \Rightarrow y^{2} - 7 y - 8 = 0 \to y = 8_{} o r y = - 1, w h i c h i s n o t p o s s i b l e \sin c e y > 0 \\ If y = 8 \to 2^{x} = 8 \\ x = 3 . \end{array}$

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Arithmetic progression

Remember concepts with our Masterclasses.

If log52,log52x−3,log5172+2x-1⁡ are in A.P then x is

2log52x−3=log52 +log5172+2x-1⇒2x−32=17+2x Let 2x=y⇒y2−7y−8=0→y=8 or y=-1, which is not possible since y>0 If y=8→2x=8x=3.

Ready to Test Your Skills?

free study material

$If \log_{5}^{2}, \log_{5}^{(2^{x} - 3)}, \log_{5}^{(\frac{17}{2} + 2^{x - 1})} are in A.P then x is$